Test: Circles- 2 - GMAT MCQ

# Test: Circles- 2 - GMAT MCQ

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## 20 Questions MCQ Test Quantitative for GMAT - Test: Circles- 2

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Test: Circles- 2 - Question 1

### The diameter of the tyres of a car is 63 centimetres. If the car is driven at a constant speed such that the tyre makes one 360-degree rotation in 0.24 seconds, what is the approximate speed of the car in kilometres per hour?

Detailed Solution for Test: Circles- 2 - Question 1

Given:

• Let the diameter of the tyre = d
• d = 63 cm
• Tyre makes one 360-degree rotation in 0.24 seconds

To find: Approx. speed of car in kilometers per hour ( in short kmph)

Approach:

1. Speed = Distance/Time
• We will find the distance that the tyre covers in 0.24 seconds
• This distance = Circumference of the tyre
• (Note: the car moves a distance through the rotation of its tyres. This is the reason why in answer to the question on the speed of the car, we are finding the distance covered by its tyres per unit time)

1. Since we’re asked for only an approximate speed, we can use the principles of Estimation and Rounding to ease our calculations. But we will not round numbers aggressively since a few answer choices are not very far apart in magnitude (Choices A and B, Choices D and E)

Working Out:

• Finding the circumference of the tyre
• Circumference of the tyre

​​​​

• Thus, the speed of the car is 30 kilometers per hour approx.

Looking at the answer choices, we see that the correct answer is Option B

Test: Circles- 2 - Question 2

### Point A lies on a circle whose center is at point C. Does point B lie inside the circle? BC2 = AC2 + AB2 ∠CAB is greater than ∠ABC

Detailed Solution for Test: Circles- 2 - Question 2

Steps 1 & 2: Understand Question and Draw Inferences

Given: The given information can be represented visually as in the figure above

To find: Does point B lie inside the circle?

• The answer will be YES if distance between point B and center C is less than the radius of the circle, that is, if BC < r

• Else, if BC = r or greater than radius r, then the answer will be NO

Step 3: Analyze Statement 1 independently

Statement 1 says that ‘BC2 = AC2 + AB2

• Now, AC = r
• The minimum possible value of AB = 0 (happens if point B lies ON point A)
• So, the minimum possible value of BC2 = r2 + 0 = r2
• Thus, the minimum possible value of BC = r
• So, BC is equal to or greater than r
• Therefore, the answer to the question is NO
• So, Statement 1 alone is sufficient to answer the question

Step 4: Analyze Statement 2 independently

Statement 2 says that ‘∠CAB is greater than ∠ABC’

• In ?ABC, since ‘∠CAB is greater than ∠ABC’, BC > AC
• (Because we know that the side opposite to the greater angle in a triangle is greater)
• Therefore, BC > r
• So, the answer to the question is NO
• Therefore, Statement 2 alone is sufficient to answer the question

Step 5: Analyze Both Statements Together (if needed)

Since we’ve already arrived at a unique answer in each of Steps 3 and 4, this step is not required

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Test: Circles- 2 - Question 3

### The tyre of a car made an average of 500 revolutions per minute in a particular trip. If the radius of the tyre was 10 centimeters and the car was driven for 1 hour without any stoppages, approximately how much distance in kilometers was covered in the trip?

Detailed Solution for Test: Circles- 2 - Question 3

Given

• Average revolutions made by a car = 500 per minute
• Radius of car = 10 centimeters
• Time for which car was driven = 1 hour

To Find: Total distance covered by the car in kilometers?

Approach

1. (Total distance covered by the car in 1 hour) = (Total distance covered by a wheel of the car in 1 hour)

2. (Total distance covered by a wheel of the car in 1 hour) = (Number of revolutions of the wheel in 1 hour) * (distance covered in each revolution, in kilometers)

3. (Number of revolutions of the wheel in 1 hour) = (revolutions per minute) * (number of minutes in 1 hour)
1. Number of minutes in 1 hour = 60 minutes

4. Distance covered in each revolution = 2πr, where r is the radius of the tyre (in cenimeters)
5. After calculating the total distance covered, we need to convert it into kilometers

Working Out

1. Distance covered in each revolution = 2πr = 20π

2. Total number of revolutions in 1 hour = 500 * 60 = 30000

3. Total distance covered in 1 hour = 30000 * 20π centimeters

4. However, we need to find the distance covered in kilometers

1. So, total distance covered in kilometers =
2. As π = 3.14, 6π ~19 kilometers

Test: Circles- 2 - Question 4

ΔABC, which is right-angled at B, is inscribed in a circle with centre O and radius 6 units. If the length of the smaller arc between points A and B is 4π units, what is the length of line segment BC?

Detailed Solution for Test: Circles- 2 - Question 4

Given:

• The figure for the given information looks like this:

• Since Triangle ABC is a right triangle inscribed in the circle, this means the hypotenuse AC of the triangle must be the diameter of the circle.
• Therefore, AC = 6*2 = 12 units
• Let the length of BC be x units.
• Length of the smaller arc between points A and B is 4π
• units

To find: x = ?

Approach and Working:

• We are given the length of the smaller arc between points A and B. Using this information, we can find ∠AOB
• Finding ∠AOB

• We’ll now use ∠AOB to find ∠BAC in isosceles triangle AOB

• Now that we know ∠BAC, we can clearly infer that triangle ABC is 30-60-90 Triangle. Using the side-angle ratio property, we can find the value of x.
• Finding x

Looking at the answer choices, we see that the correct answer is Option D.

Test: Circles- 2 - Question 5

In the figure above, triangle ABC is inscribed in the circle with center O, such that CD is perpendicular to AB. If the length of the side AC is centimeters, ∠AOB is 60o  and the smaller perimeter of the sector AOB is 12 + 2π centimeters, what is the perimeter of the triangle ABC in centimeters?

Detailed Solution for Test: Circles- 2 - Question 5

Given

• Circle with center at O
• Let the radius of the circle be r
• Triangle ABC is inscribed in the circle
• CD is perpendicular to AB
• So, OD is also perpendicular to AB. Hence AD = DB
• This is because the perpendicular drawn from the centre of the circle to the chord of the circle bisects the chord.
• AC =
• centimeters
• Perimeter of minor sector AOB = 12 +2π centimeters
• In triangle AOB,
• ∠AOB = 60o
• Triangle AOB is an isosceles triangle, with OA =OB =r
• So, ∠OAB = ∠OBA
• Now, since ∠AOB = 60o
• ⇒ ∠OAB = ∠OBA = ∠AOB = 60o
• Triangle AOB is an equilateral triangle
• So, OA =OB = BA =r (All sides are equal in an equilateral triangle)
• To Find: Perimeter of triangle ABC

• Perimeter of triangle ABC = AC + CB + BA
• We know AC =  cm
• So, Perimeter of triangle ABC =+ CB+BA
• So, to answer the question, we need to find the length of CB and BA.

Approach

• Finding length of CB
1. In triangle ABC, we have the perpendicular from vertex C bisecting the opposite base AB, hence triangle ABC should be an isosceles triangle with AC = CB.
2. As we know the length of AC, we can find the length of CB

• Finding length of BA
• Length of BA is equal to r
• We know that perimeter of sector AOB =
1. OA = OB = r (already established above)
2. We are also given perimeter is equal to 12 +2π
3. So from the above relation we will be able to calculate the value of r. which is equal to BA.
• Knowing the value of CB and BA, we will be able to find the perimeter of triangle ABC
• Working Out

• Finding length of CB
1. We know that AC =centimetres and AC = CB.
2. Thus CB = centimetres
3. Finding length of BA

hence BA = 6 centimeters

Test: Circles- 2 - Question 6

In the given figure, AB and CD are the longest chords of the circle with their lengths equal to 8 units. If the length of the minor arc BC is 1/6th of the perimeter of the circle, what is the length of the chord BD?

Detailed Solution for Test: Circles- 2 - Question 6

Given

• AB = CD = 8 units and they are the longest chords of the circle.
• AB and CD are diameters of the circle.
• As diameters are the longest chord.
• Let their intersection point be O, center of the circle.
• Radius of the circle = 4 units
• Length of minor BC = 1/6∗(Perimeter of the circle)
• = 1/6∗2πr

To Find: Length of chord BD?

Approach

• We need to find the length of chord BD. We notice that chord BD is a part of triangle OBD.
1. In triangle OBD, we know that OB = OD = 4 units (radius of the circle)
2. The third side of this triangle, BD, is unknown that we need to find out.
• Now, since OBD is an isosceles triangle with OB=OD, if we drop a perpendicular in this triangle it will bisect the BD.
1.  Let’s drop a perpendicular from O to chord BD at point E.
1. Therefore, BE = ED

Now as triangle OED is a right angled triangle, knowing one of the sides and one of the angles will be sufficient to calculate the length of ED.

• In right triangle ODE,
1. We know that OD = 4 units.
1. So, we need to find one of the angles of triangle ODE.
2. As triangle OBD is isosceles and OE is perpendicular from O, OE will bisect ∠BOD. So, ∠DOE = ∠BOD/2
3. Also, ∠BOD = 180  - ∠COB. So, if we can calculate ∠COB, we can find the value of ∠BOD
• We know that length of arc BC =
• We can use the above relation to calculate the value of ∠COB

Working Out

1. Length of arc  which gives us ∠COB = 60o

4. As triangle ODE is a 30o - 60o -90o  triangle, we have the length of ED = 2√3

5. Thus BD = 2 * 2√3 = 4√3

Test: Circles- 2 - Question 7

If O is the center of the circle above and the length of chord AB is 2 units, what is the length of the arc ACB?

(1) The area of ΔOAB is √3 square units

(2) The area of sector OACBO is 2π/3

Detailed Solution for Test: Circles- 2 - Question 7

Steps 1 & 2: Understand Question and Draw Inferences

Given:

• AO = OB = r, the radius of circle
• AB = 2

Step 3: Analyze Statement 1 independently

(1) Statement 1 states that "The area of ΔOAB is √3 square units."

• Let’s drop a perpendicular OD from O to AB. Let the length of this perpendicular be h

• Also, AD = 1 (since perpendicular OD will bisect the chord AB)
• So, in right triangle ODA,

r=2 (Rejecting the negative root since radius cannot be negative))

• Since we now know the value of r and ∠ AOB, we can find the length of arc.
• Hence, Statement 1 is sufficient to answer the question.

Step 4: Analyze Statement 2 independently

(2) Statement 2 states that "The area of sector OACBO is 2π/3

• Multiple combinations of r and ∠ AOB will satisfy this equation.
• Example, r = 1 and ∠ AOB = 120o
• Or r = 2 and
• Thus, Statement 2 is not sufficient to answer the question.

Step 5: Analyze Both Statements Together (if needed)

Since we’ve already arrived at a unique answer in Step 3, this step is not required

Hence the correct answer is Option A .

Test: Circles- 2 - Question 8

The annual money spent by a company on search engine marketing, content marketing and affiliate marketing is to be shown on a circlular graph. If the size of each sector of the graph is proportional to the amount of money it represents, did the company spend the most money on search engine marketing?

(1) The angle of the sector that represents content marketing is 80o

(2) The angle of the sector that represents affiliate marketing is 210o

Detailed Solution for Test: Circles- 2 - Question 8

Steps 1 & 2: Understand Question and Draw Inferences

Given: 3 amounts to be represented on a circle graph:

• Search Engine Marketing Spending
• Content Marketing Spending
• Affiliate Marketing Spending
• Let the angles of the respective sectors be s∘,c∘,a∘
•  respectively

To find: Was money spent on Search Engine Marketing the greatest?

• The answer will be YES if, out of s,c,a,s
•  is the greatest
• Now, we know that  s+c+a=360

Step 3: Analyze Statement 1 independently

Statement 1 says that ‘The angle of the sector that represents content marketing is 80∘

• Based on the above equation, we cannot say if s∘ is the greatest angle or not
• For example, if s=220∘ and a ∘=60 then the answer is YES
• But if s=60 and a=220 then the answer is NO
• So, Statement 1 is not sufficient to arrive at a unique answer

Step 4: Analyze Statement 2 independently

Statement 2 says that ‘The angle of the sector that represents affiliate marketing is 210∘

• Now, from the above equation, we can infer that the greatest possible value of s∘is150∘ (happens when c∘=0
• So, even the greatest possible value of s is less than the measure of a∘ (which is 210)
• Therefore, s is definitely not the greatest of the 3 angles
• So, the answer to the question is NO
• Since we’ve been able to arrive at a unique answer, Statement 2 alone is sufficient.

Step 5: Analyze Both Statements Together (if needed)

Since we’ve already arrived at a unique answer in Step 4, this step is not required

Test: Circles- 2 - Question 9

The figure above shows a rectangular plot of land OPQR. In this plot, a patch of land OABO, where the path AB represents a circular arc of radius 20√2 metres, is developed as a lawn. The point A is at a distance of 20 metres each from sides OR and OP. The point B is at a distance of 10√6 metres from side OP and 10√2 metres from side OR. What is the area, in square metres, of the intended lawn?

Detailed Solution for Test: Circles- 2 - Question 9

Given:

• Let’s consider OP as the x-axis and OR as the y-axis.
• So, the coordinates of points A and B are: A(20, 20) and B(10√2, 10√6)

To find: Area of region OABO

Approach:

1. We’re given that AB is a circular arc of radius 20√2 meters but we are not told that region OABO represents a sector of a circle of radius 20√2 meters. So, it will be wrong to assume that OA and OB are the radii of the sector OABO. In fact, we’ll need to find the values of OA and OB to check if they are indeed the radii of circular arc AB.
1. We can find the value of OA and OB using the Pythagoras theorem in triangles OEA and OFB.
2. If OA and OB are indeed the radii of the circular arc, then we can use the formula of area of sector to find the area of region OABO. But to find this area, we’ll also need the measure of ∠BOA. We can find this angle as follows
• ∠BOA = ∠BOF - ∠AOE
• ∠BOF can be found by applying trigonometric ratios in ΔOFB

Working Out:

Thus, we see that OA = OB = Radius of the circular arc AB =20√2  meters

• This means that OABO is indeed the sector of a circle of radius 20√2 meters
• So, we’ll be able to apply the formula for area of a sector to answer the question.

Thus, ∠BOA =  15o

Finding the area of sector OABO

• So, the area of sector

Looking at the answer choices, we see that the correct answer is Option B

Test: Circles- 2 - Question 10

A circle is inscribed in a square ABCD such that the circle touches all the sides of the square. If the perimeter of the shaded region is 24 + 6π, what is the area of the shaded region?

Detailed Solution for Test: Circles- 2 - Question 10

Given

• Square ABCD

A circle is inscribed in the square ABCD

Let the side of square be x and the radius of the circle be r

• Circle touches all sides of the square, so diameter = side of square
• 2r = x
• Perimeter of the shaded region = 24 + 6π

To Find: Area of the shaded region?

Area of the shaded region = (3over 4)(Area of square – Area of circle)

1. We can observe that the area in the figure that is outside the circle but inside the square are four parts as shown in the above figure.
2. Area of (Part A + Part B + Part C + Part D) = Area of square – Area of Circle.

Out of these 4 parts:

1. Part A, B & C are shaded.
3. Now the figure is symmetric about the centre axis, so each part of the 4 parts will have same side length.
1. Thus, all the parts i.e. Part A, B, C & D will also be equal in area.
4. Therefore, Area of Part A = Area of Part B = Area of Part C = Area of Part D
1. 4*Area of each part = Area of square – Area of Circle
2. Area of each part = 1/4 (Area of square - Area of circle)
• We need to find the areas of only 3 shaded parts, thus area of the shaded region = 3/4(Area of square – Area of circle)
1. Thus to find the area of the shaded region, we need to find the values of Area of Square and Area of Circle.

Approach

3. Since area of square and area of circle are in terms of r, we need to find the value of r, by equating perimeter of shaded region to 24 + 6π.

Perimeter of the shaded region = Perimeter of the square + Perimeter of the circle – perimeter of the region AFE

1. Perimeter of square = 4x = 8r
2. Perimeter of circle = 2πr = πx
3. Calculating Perimeter of the region AFE:
1. If, we join OF and OE, we have AEOF as a square with side length = r = x2
•
• So, we have the length of AF and AE in terms of r, also he angle made by arc FE on center O(90o)(due to symmetry of figure about the centre axes), we can calculate the length of arc FE using the formula
• Thus we can calculate the perimeter of region AFE in terms of x, where r = x/2
• .
• So perimeter of region AFE =

Working Out

1. Perimeter of the shaded region = Perimeter of the square + Perimeter of the circle – perimeter of the region AFE
• On solving w, we get x = 8

2. Area of shaded region = 3/4 ∗(Area of square – Area of circle)

• Therefore, the correct answer is Option C.
Test: Circles- 2 - Question 11

Two concentric circles have their centers at point O such that a line segment AB having its end points on the outer circle touches the inner circle at point C. The length of the line segment AB is times the radius of the inner circle. If an equilateral triangle is drawn such that the area of the triangle is equal to the ratio of the radii of the outer circle and the inner circle respectively, what is the length of the side of the triangle?

Detailed Solution for Test: Circles- 2 - Question 11

Given

• Two concentric circles with center at O
• Let’s assume the radius of the outer circle to be R and that of inner circle to be r
• AB is a chord to the outer circle and a tangent to the inner circle
•
• An equilateral triangle with Area = R/r
• Assuming the side length of the equilateral triangle to be a, we have

To Find:

• Value of a.

Approach

• Since  to find the value of a, we need to find the value of  R/r
• We know that AB is a tangent to the inner circle, hence AB will make an angle of 90o with the radius of the inner circle, i.e. the line joining the center O to the point of tangency. Let’s call the point of tangency as C.
• So, OC will be perpendicular to AB.

• However, as AB is also a chord to the external circle, the line perpendicular from the center O will bisect the chord, i.e. OC will bisect the chord AB.
• Therefore,
• As we know that BC =​  and triangle OCB is a right angled triangle, we can use Pythagoras theorem in triangle OCB to calculate the ratio of R/r

Working Out

1. Using Pythagoras theorem in triangle OCB, we have

(rejecting the negative root since the ratio of radii cannot be negative)

2.

Therefore, a=2   , as a being the length of a triangle, cannot be negative.

Hence the correct answer is Option C .

Test: Circles- 2 - Question 12

If a circle has the same area as right triangle PQR shown in the figure above. Which of the following is the closest in value to the radius of the circle?

Detailed Solution for Test: Circles- 2 - Question 12

Given:

• Right ?PQR
• QR = 6 units
• PR = 10 units
• A circle whose area is equal to area of triangle PQR
• Let the radius of the circle = r units

To find: r = ?

Approach:

b. Therefore, to find r, we need to find A

Working Out:

c. The value of will be very slightly greater than 1

d. We know that Since 7 is slightly greater than the mid-point between 4 and 9 (which is

the value of √7 will be slightly greater than the mid-point between   So, the value of √7 will be slightly greater than 2.5

e. So, the product of   and  √7 will be slightly greater than 2.5

f. Therefore, the closest integer to the value of r will be 3

Test: Circles- 2 - Question 13

In the figure above, triangle ABC is inscribed in a circle whose centre O has the x- and y-coordinates as (0,0). If the x- and y- coordinates of point A are (-4,0) and ∠BAC = 30, what is the area, in square units, of triangle AOB?

Detailed Solution for Test: Circles- 2 - Question 13

Given:

O is the center of the circle and A(-4,0) is a point on the circle

• So, radius of the circle, R = Distance between points O and A = 4 units
• Also, AC is the diameter of the circle.

S​o, ∠ABC is an angle in the semicircle

• So, ∠ABC = 90o
• In right ΔABC,
• ∠BAC = 30∘

So, by Angle Sum Property, ∠ACB = 60o

To find:

• Area of (ΔAOB)

Approach:

• Let’s drop a perpendicular BD from point B on AC.

Area triangle AOB = ½ * AO * BD

• We’ve inferred the length of AO (= radius of the circle) in the Given section above. So, to answer the question, we need to find the length of BD
• In ΔOBC, sides OB and OC are equal because both are the radii of the circle.

So, this is an isosceles triangle.

• Since we already know ∠OCB=60, therefore ∠OBC will also be equal to 60.
• Which finally helps us in inferring that ∠BOC = 60o
• Hence we can conclude that OBC is an equilateral triangle.
• The ΔODB, is a 30-60-90 Triangle and we know the measure of radius OB = 4 units. Thus using the side property of 30-60-90 Triangle we can write –

OD: BD: OB = 1: √3: 2

• BD: OB = √3: 2
• Thus BD = 2√3
• Thus area of triangle AOB = ½ * AO * BD = ½ * 4 * 2√3 = 4√3
• Looking at the answer choices, we see that the correct answer is Option B.

Test: Circles- 2 - Question 14

Lines CA and CB touch the circle only at points A and B respectively, as shown in the figure above. If the centre O of the circle is at the point (0,0), the coordinates of point C are (0,2) and the radius of the circle is √2 units, what are the coordinates of point A?

Detailed Solution for Test: Circles- 2 - Question 14

Given:

• CA and CB are tangents to the circle
• Since tangents from the same point are equal, CA = CB
• Also, ∠OBC = ∠OAC = 90o
• Center of the circle is O (0,0) and Radius = √2 units
• Point C(0,2)

To find:

• The x- and y- coordinates of Point A

Approach and Working:

• Let’s drop a perpendicular AD from A on the x-axis

• So, the x-coordinate of point A = OD
• And, the y-coordinate of point A = AD

• So, in order to answer the question, we need to find OD and AD
• In right ΔAOC, we know sides OA = √2 and OC = 2 units
• Using Pythagoras Theorem, we can write –
• OC2 = OA2 + AC2
• AC2 = OC2 – OA2 = 4 – 2 = 2
• Hence AC = √2
• Thus, we can conclude that ΔAOC is an isosceles right-angled triangle, where angle ACO = AOC = 45o.
• Angle AOD = 90o – angle AOC = 900 – 45o = 45o.
•

• In triangle AOD, angle AOD = 45, and angle ADO = 90, thus angle DOA = 45
• Hence Triangle AOD is a 45-45-90 Triangle in which we can write the side ratio as –
• AD: OD: AO = 1: 1: √2
• Since the exact value of AO = √2
• We can conclude from the ratio above, that AD = OD = 1 units
• Thus, we see that the coordinates of point A are (1,1)
• Looking at the answer choices, we see that the correct answer is Option C .
Test: Circles- 2 - Question 15

In the coordinate system, the center of a circle lies at (2, 3). If point A with coordinates (-1, 7) does not lie outside the circle, which of the following points must lie inside the circle?

I. (0, 7)

II. (5, -1)

III. (-2, 7)

Detailed Solution for Test: Circles- 2 - Question 15

Given

• Circle with center at (2, 3)
• Let’s assume the radius of the circle to be r.
• Point A(-1, 7) does not lie outside the circle

To Find: Which of the points in the options must lie inside the circle?

Approach

1. To know the points that must line inside the circle, we need to find the distance between the center of the circle and the point (x, y)

• Let us understand how can we find the distance between 2 points in the coordinate plance.
• Let us say the above figure represent the coordinates of one point → point A → (a,b) and of another point → Point C → (x,y) in the co-ordinate system
1. Distance of point A from point B along the x-axis = (x-a)
1. So, AB=x-a
2. Distance of point C from point B along the y-axis = (y-b)
1. So, CB= y-b
3. Now, the triangle ABC formed is a right angled triangle, so by Pythagoras theorem

In general, it can be said that the distance between any two points is =

So we can find the distance between Point A and the centre of the circle by the above formula

• If this distance between the center of the circle and point(x, y) is less than the minimum possible value of r, then the point (x, y) must lie inside the circle.
• So, we need to find the minimum possible value of r.
• We are given that point A does not lie outside the circle. So, it may lie either inside the circle or on the circle.
1. Hence (the distance of point A from the center) ≤ r
1. That is, r ≥ (the distance of point A from the center)
1. So, (minimum possible value of r) = (the distance of point A from the center)
1. We will use the above relation to find the minimum possible value of r.

Working Out

1. Coordinates of point A = ( -1, 7) and coordinates of center of the circle = (2, 3)
1. Distance of point A from the center of the circle ​
2. Thus the minimum possible value of r = 5
2. Evaluating the 3 options
1. Distance of point (0, 7) from the center (2, 3)
1. ​​​
3. As the distance of point is less than the minimum possible value of r, this point must lie inside the circle
1. Distance of point (5, -1) from the center (2, 3)
1. As the distance of point is not less than the minimum possible value of r, this point may or may not lie inside the circle.
1. For example, if r > 5, the point will lie inside the circle
2. If r = 5, the point will not lie inside the circle, it will lie on the circle.
2. Distance of point (-2, 7) from the center (2, 3)
1. As the distance of point is not less than the minimum possible value of r, this point may or may not lie inside the circle

Hence, we see that only option I (0, 7) always lie inside the circle.

Test: Circles- 2 - Question 16

15 children are given tags numbered from 1 to 15 and are seated in a circular formation in the increasing order of their respective tag numbers. The total area covered by the circular formation is 36π square units and the distance between any two neighbouring children in the formation is equal., If the number of children seated between the child with tag number m and the child with tag number 1 is equal to the number of children seated between the child with tag number m and the child with tag number 15, what is the minimum distance covered along the circular formation by the child with tag number 1 to reach the child with tag number m -2 and then go back to his original position?

Detailed Solution for Test: Circles- 2 - Question 16

Given

• 15 children are seated at equal distance in a circular formation
• Let the distance between two neighboring students be x.

• Area of the circular formation = 36π
• Let’s assume the radius of the circular formation be r
• πr2=36π
• Child number ‘m’ is equidistant from child number 1 and child number 15

To Find:

• Minimum distance covered by child number 1 to reach child number m -2 and back to his original position

Approach

• Let’s understand first how we can find the distance between any two children.
• Let’s assume we need to find the distance between child number 1 and child number 4.
• As the distance between two neighboring children is x, distance between child number 4 and child number 1 will be
1. Case 1: Distance =(4-1) * x = 3x
2. Case 2: Distance = 15x – 3x = 12x
1. 15x is the total distance along the circular formation for 15 children.
3. So minimum distance will be = minimum {distance a, distance b} = minimum {3x,12x} = 3x
• So based on the example above we can infer that in general the minimum distance between any two children = Minimum {distance c, distance d}, where
1. distance c = ((Position of first child – Position of second child))*x
2. distance d = 15x – ((Position of first child – Position of second child))*x
• Now, distance c & distance d or any distance between children is a part of the perimeter of the circle.
1. So, we can write 15x = 2πr
• Thus, if we know the value of r, we can find the value of x.

• Hence, for finding the distance traveled, we need to find the following:
1. Value of m
2. Radius of the circle i.e. r

• Value of m
1. As child m is equidistant from child number 1 and child number 15, we can write
1. Distance between child number m and child number 1 = Distance between child number 15 and child number m
2. (m-1)*x = (15-m)*x
3. This will give us the value of m and hence we can find the value of m -2

1. We know that πr2=36π
2. We can use the above equation to find out the value of r.

Working Out

• Finding value of m
• (m – 1)*x = (15- m)*x, i.e. m = 8
• Hence, m – 2 = 6
• Calculating minimum distance
• distance c = ((position of child number m - 2 ) – (position of child number 1))*x
• So, distance c = (6-1)*x = 5x
• distance d = 15x - ((position of child number m - 2 ) – (position of child number 1))*x
• So, distance d = 15x – 5x
• Minimum distance = minimum {distance c, distance d}
• So, minimum distance = minimum {5x,10x} = 5x
• Distance to go back to original position = 5x+5x =10x

• Therefore, the correct answer is Option C.
Test: Circles- 2 - Question 17

The figure above shows a circle whose centre O has the x- and y- coordinates as (6,0). Points A(m,n) and B(8,2) are marked on the circle such that ∠AOB = 105. If the circle is symmetrical about the x-axis, what is the value of m?

Detailed Solution for Test: Circles- 2 - Question 17

Given:

• Circle
• Center = O(6,0)
• Points on the circle: A(m,b) and B(8,2)
• This means, OA = OB = radius of circle
• ∠AOB = 105∘

To find: m = ?

Approach:

1. Let’s drop a perpendicular from point A on the x-axis.

From this diagram, it’s clear that there are 2 ways to find the value of m:

1. Way 1 – In right triangle AFO, if we can know the measure of one angle (out of ∠AOF and ∠OAF) and one side (like the radius AO), we can use trigonometric ratios to find length OF, and hence the value of m (since OF = 6 – m)
2. Way 2 – In right triangle AFO, if we can know the measure of sides AO (radius of circle) and AF, we can use Pythagoras Theorem to find OF, and hence m.

2. We are given the coordinates of both points O and B. So, using these, we can find the radius OB = OA

3. Also, if we drop a perpendicular from point B on the x-axis, we’ll know the magnitude of all the sides of the resulting right triangle.

So, we can use trigonometric ratios to find ∠BOE.

So, we can find ∠AOF using the equation

(Thus, in this Approach, we’re using Way 1 to find m)

Working Out:

• Radius OB = OA =  2√2

Looking at the answer choices, we see that the correct answer is Option D

Test: Circles- 2 - Question 18

If O is the center of the circle above, what fraction of the circular region is encompassed by an angle of degrees?

Detailed Solution for Test: Circles- 2 - Question 18

In this question, the angles shown appear span half the circle. We can infer that they must, in fact, because the opposite angles of bisecting lines are equal. On the other side of each of the 2x angles shown is another 2x, and on the other side of each x shown is another x. Therefore,
2x+x+2x+x = 180

half the span of the circle. Thus, and . Since the full span of a circle is 360 degrees, the angle x spans a fraction of 30/360 = 1/12

Test: Circles- 2 - Question 19

In the coordinate plane, a circle has center (-1, -3) and passes through the point (4,2). What is the circumference of the circle?

Detailed Solution for Test: Circles- 2 - Question 19

To find the circumference of this circle, we need its radius. And the radius is the distance from its center to any point on the circle, so the radius is the distance from (-1 ,-3) to (4,2).

This distance is computed with the distance formula, which is essentially the Pythagorean Theorem. The sides of the triangle are the x and y distances, 5 and 5, so the distance is
The radius of the circle is 5√2, so the circumference is

Test: Circles- 2 - Question 20

The circle with center C shown above is tangent to both axes and has an area of A. What is the distance from O to C, in terms of A?

Detailed Solution for Test: Circles- 2 - Question 20

In this question, we can find the radius r in terms of A by manipulating the area formula, A= πr2:

Point C in the figure is the upper-right point of a square that has sides of length r. And the distance from the origin to C is the diagonal of the square, so it is the hypotenuse of a right triangle with height r and base r, so the distance will equal

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## Quantitative for GMAT

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