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All questions of Circle for JEE Exam

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 The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Poonam Reddy answered
 Radius of circle is given by -
r = √[(h-x1)² + (k-y1)²]
r = √[(2-3)² + (1+5)²]
r = √(-1² + 6²)
r = √(1 + 36)
r = √37
if centre (2,-]1) and radius=√26 are given,
(x-h)2+(y-k)2=r2
equation is (x-2)2 + (y-1)2 = (√37)2
x2 + 4 - 4x + y2 + 1 - 2y = 37
x2 + y2 - 4x - 2y - 32 = 0

The equation of the circle passing through (0, 0) and making intercepts 2 and 4 on the coordinate axes is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Geetika Shah answered
The circle intercept the co-ordinate axes at a and b. it means x - intercept at ( a, 0) and y-intercept at (0, b) .
Now, we observed that circle passes through points (0, 0) , (a, 0) and (0, b) .
we also know, General equation of circle is
x² + y² + 2gx + 2fy + C = 0
when point (0,0)
(0)² + (0)² + 2g(0) + 2f(0) + C = 0
0 + 0 + 0 + 0 + C = 0
C = 0 -------(1)
when point (a,0)
(a)² + (0)² + 2g(a) + 2f(0) + C = 0
a² + 2ag + C = 0
from equation (1)
a² + 2ag = 0
a(a + 2g) = 0
g = -a/2
when point ( 0, b)
(0)² + (b)² + 2g(0) + 2f(b) + C = 0
b² + 2fb + C = 0
f = -b/2
Now, equation of circle is
x² + y² + 2x(-a/2) + 2y(-b/2) + 0 = 0 { after putting values of g, f and C }
x² + y² - ax - by = 0
As we know that, a=2, b=4
x^2 + y^2 - 2x - 4y = 0

 The centre and radius of the circle x2 + y2 + 4x – 6y = 5 is:
  • a)
    (2, – 3), 2√2
  • b)
    (– 2, 3), 3√2
  • c)
    (– 2, 3), 2√2
  • d)
    (2, – 3), 3√2
Correct answer is option 'B'. Can you explain this answer?

Riya Banerjee answered
x2+y2+4x-6y=5
Circle Equation
(x-a)2+(y-b)2=r2 is the circle equation with a radius r, centered at (a,b)
Rewrite x2+y24x-6y=5 in the form of circle standard circle equation
(x-(-2))
2
+(y-3)2=(3
√2)2
Therefore the circle properties are:
(a,b) = (-2,3), r = 3√2

The equation of the circle having the lines y2 – 2y + 4x – 2xy = 0 as its normals & passing through the point (2, 1) is
  • a)
    x2 + y2 – 2x – 4y + 3 = 0
  • b)
    x2 + y2 – 2x + 4y – 5 = 0
  • c)
    x2 + y2 + 2x + 4y – 13 = 0
  • d)
    None
Correct answer is option 'A'. Can you explain this answer?

Geetika Shah answered
The normal line to circle is →y² - 2 y + 4 x -2 xy=0
→ y(y-2) - 2x(y-2)=0
→ (y-2)(y-2x)=0
the two lines are , y=2 and 2 x -y =0
The point of intersection of normals are centre of circle.
→ Put , y=2 in 2 x -y=0, we get
→2 x -2=0
→2 x=2
→ x=1
So, the point of intersection of normals is (1,2) which is the center of circle.
Also, the circle passes through (2,1).
Radius of circle is given by distance formula = [(1-2)² + (2-1)²]½ 
=(1+1)½ =(2)½
The equation of circle having center (1,2) and radius √2 is
= (x-1)²+(y-2)²=[√2]²
→ (x-1)²+(y-2)²= 2
x²+y² -2x-4y+3 = 0

The equations of the tangents drawn from the point (0, 1) to the circle x2 + y2 - 2x + 4y = 0 are
  • a)
    2x - y + 1 = 0, x + 2y - 2 = 0
  • b)
    2x - y - 1 = 0, x + 2y - 2 = 0
  • c)
    2x - y + 1 = 0, x + 2y + 2 = 0
  • d)
    2x - y - 1 = 0, x + 2y + 2 = 0
Correct answer is option 'A'. Can you explain this answer?

Geetika Shah answered
Let equation of tangent with slope =m and point (0,1)
(y−1)=m(x−0)⇒y=mx+1
Intersection point
x2+(mx+1)2−2x+4(mx+1)=0
(1+m2)x2+(−2+6m)x+5=0
For y=mx+1 to be tangent, discriminant =0
(6m−2)2−4×5(1+m2)=0
36m2+4−24m−20m2+20=0
16m2−20m+24=0
⇒ 2m2−3m−2=0
(2m+1)(m−2)=0

The locus of the mid-points of the chords of the circle x2 + y2 – 2x – 4y – 11 = 0 which subtend 60º at the centre is
  • a)
    x2 + y2 – 4x – 2y – 7 = 0
  • b)
    x2 + y2 + 4x + 2y – 7 = 0
  • c)
    x2 + y2 – 2x – 4y – 7 = 0
  • d)
    x2 + y2 + 2x + 4y + 7 = 0
Correct answer is option 'C'. Can you explain this answer?

Preeti Iyer answered
Let AB be the chord of the circle and P be the midpoint of AB.
It is known that perpendicular from the center bisects a chord.
Thus △ACP is a right-angled triangle.
Now AC=BC= radius.
The equation of the give circle can be written as
(x−1)2+(y−2)2=16
Hence, centre C=(1,2) and radius =r=4 units.
PC=ACsin60degree
= rsin60degree
= 4([2(3)½]/2
= 2(3)1/2 units
Therefore, PC=2(3)1/2
⇒ PC2=12
⇒ (x−1)2+(y−2)2=12
⇒ x2+y2−2x−4y+5=12
⇒ x2+y2−2x−4y−7=0

y = √3x + c1 & y = √3x + c2 are two parallel tangents of a circle of radius 2 units, then |c1 – c2| is equal to
  • a)
    8
  • b)
    4
  • c)
    2
  • d)
    1
Correct answer is option 'A'. Can you explain this answer?

Anmol Chauhan answered
For both lines to be parallel tangent the distance between both lines
should be equal to the diameter of the circle
⇒ 4 = |c1−c2|/(1+3)1/2
⇒∣c1−c2∣ = 8

 Pair of tangents are drawn from every point on the line 3x + 4y = 12 on the circle x2+ y2 = 4. Their variable chord of contact always passes through a fixed point whose co-ordinates are
  • a)
  • b)
  • c)
    (1, 1)
  • d)
     
Correct answer is option 'D'. Can you explain this answer?

Geetika Shah answered
Let P(x1,y1) be a point on the line 3x + 4y = 12
Equation of variable chord of contact of P(x1,y1) w.r.t circle x2 + y2 = 4 
xx1 + yy1 − 4 = 0   ...(1)
Also 3x1 + 4y1 − 12 = 0
⇒ x1 + 4/3y1 − 4 = 0   ...(2)
Comparing (1) and (2), we get
x = 1; y = 4/3
∴ Variable chord of contact always passes through (1, 4/3)

If a be the radius of a circle which touches x-axis at the origin, then its equation is
  • a)
    x2 + y2 + ax = 0
  • b)
    x2 + y2 ± 2ya = 0
  • c)
    x2 + y2 ± 2xa = 0
  • d)
    x2 + y2 + ya = 0
Correct answer is option 'B'. Can you explain this answer?

Sushant Ghosh answered
The equation of the circle  with centre at (h,k) and radius equal to a is (x−h)2 +(y−k)2 = a2
 When the circle passes through the origin  and centre lies on x− axis 
⇒h = a and k = 0
Then the equation (x−h)2+(y−k)2=a2 becomes (x−a)2+y2=a2
If a circle passes through the origin and centre lies on x−axis then the abscissa will be equal to the radius of the circle and the y−co-ordinate of the centre will be zero Hence, the equation of the circle will be of the form 
(x±a)2+y2=a2⇒x2+a2 ±2ax+y2=a2
=x2 +y2 ±2ax=0 is the required equation of the circle.

If the line 2x – y + λ = 0 is a diameter of the circle x2+y2+6x−6y+5 = 0 then λ =
  • a)
    6
  • b)
    9
  • c)
    3
  • d)
    12
Correct answer is option 'B'. Can you explain this answer?

Rajesh Gupta answered
x2 + y2 + 6x − 6y + 5 = 0
Center O = (-3, 3)
radius r = √{(-3)2 + (3)2 - 5} 
= √{9 + 9 - 5} 
= √13
Since diameter of the circle passes through the center of the circle.
So (-3, 3) satisfies the equation 2x – y + λ = 0
=> -3*2 - 3 + λ = 0
=> -6 - 3 + λ = 0
=> -9 + λ = 0
=> λ = 9

The equation of a circle which passes through the three points (3, 0) (1, –6), (4, –1) is
  • a)
    2x2 + 2y2 + 5x – 11y + 3 = 0
  • b)
    x2 + y2 – 5x + 11y – 3 = 0
  • c)
    x2 + y2 + 5x – 11y + 3 = 0
  • d)
    2x2 + 2y2 – 5x + 11y – 3 = 0
Correct answer is option 'D'. Can you explain this answer?

Pooja Shah answered
Gen equation of a circle is (x−h)2+(y−k)2=r2 (3,0),(1,−6),(4,−1) circle passes through these points
∴ These three points should satisfy the equation
of circle.
(3−h)2+k2=r2 ___ (I)
(1−h)2 +(−6−k)2 = r2(1−h)2+(6+k)2=r2  ___ (II)
From (I) & (II)
(3−h)2+k2 =(1−h)2+(6+k)2
9+h2−6h+k2 =1+h2 2h+36+k2+12k
9−6h=1−2h+36+12k
9−37=4h+12k
−28=4h+12k
h+3k=−7
h=−7−3k ___ (III)
(4−h)2+(−1−k)2=r2 ___ (IV)
From (I) & (IV)
(3−h)2+k2 = (4−h)2 +(1+k)2
9+h2 −6h+k2
 =16+h2 −8h+1+k2+2k
9−6h=17−8h+2k
9−17=−2h+2k
−8=−2h+2k
−h+k=−4
k=−4+h ___ (V)
Put (V) in (III)
h=−7−3(−4+h)
h=−7+12−3h
h=5/4,k=−11/4,r=170/16
Eq of circle: (x−5/4)2 +(y+11/4)2 =170/16
x2+25/16−10x/4+y2 +121/16+22y/4=170/16
Simplifying, we get 2x2+xy2−5x+11y−3=0

 
 The locus of the centres of the circles such that the point (2, 3) is the mid point of the chord
5x + 2y = 16 is
  • a)
     2x – 5y + 11 = 0
  • b)
    2x + 5y – 11 = 0
  • c)
    2x + 5y + 11 = 0
  • d)
    None
Correct answer is option 'A'. Can you explain this answer?

Nandini Patel answered
Let locus lie on
Y = mx +C
Slope of a line = -5/2
Hence slope of its perpendicular line = 2/5
Hence
y = 2x/5  +C
5y = 2x + 5C
(2, 3) satisfies this
C = 11/5
Hence
Line will be
5y = 2x + 5*11/5
2x - 5y + 11 = 0

The equation of parabola whose focus is (– 3, 0) and directrix x + 5 = 0 is:
  • a)
    y2 = 4(x + 1)
  • b)
    y2 = 4x
  • c)
    y2 = 4(x + 16)
  • d)
    y2 = 4(x + 4)
Correct answer is option 'D'. Can you explain this answer?

Gaurav Kumar answered
Given, focus : (-3,0)
directrix : x + 5 = 0
Let (x ,y) is the point on the parabola .
∴ distance of point from focus = distance of point from directrix
⇒ √{(x + 3)² + y²} = |x + 5|/√(1² + 0²)
⇒ √{(x + 3)² + y² } = |x + 5|
squaring both sides,
(x + 3)² + y² = (x + 5)²
⇒y² = (x + 5)² - (x + 3)²
⇒y² = (x + 5 - x - 3)(x + 5 + x + 3)
⇒y² = 2(2x + 8) = 4(x + 4)
Hence, equation of parabola is y² = 4(x + 4)

The greatest distance of the point P(10, 7) from the circle x2 + y2 – 4x – 2y – 20 = 0 is
  • a)
    5
  • b)
    15
  • c)
    10
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Sankar Mehta answered
The equation of the circle is x^2 + y^2 = r^2, where r is the radius of the circle. We need to find the value of r first.

The circle passes through the origin (0,0), since x^2 + y^2 = 0^2 when x = 0 and y = 0. Therefore, the distance from the center of the circle to the origin is equal to the radius.

The center of the circle is at (0,0), so the distance from the center to the point P(10,7) is:

d = sqrt((10-0)^2 + (7-0)^2) = sqrt(149)

Therefore, the radius of the circle is r = sqrt(149).

The greatest distance from the point P(10,7) to the circle is the distance from the point to the edge of the circle along a line that passes through the center of the circle. This is equal to the difference between the distance from the center to the point and the radius of the circle.

The distance from the center of the circle to the point P(10,7) is d = sqrt((10-0)^2 + (7-0)^2) = sqrt(149).

Therefore, the greatest distance of the point P(10,7) from the circle is:

d - r = sqrt(149) - sqrt(149) = 0.

Therefore, the greatest distance is 0, which means that the point P(10,7) is on the circle.

The number of points on X-axis which are at a distance c units (c < 3) from (2, 3) is
  • a)
    1
  • b)
    0
  • c)
    3
  • d)
    2
Correct answer is option 'B'. Can you explain this answer?

Mohit Rajpoot answered
Distance of 'c' units from (2,3)
Let the no: of points be (x,0)
By distance formula
{(2−x)2+(3−0)2}=c
 4−4x+x2+9=c
⇒x2−4x+13 = c:c=2,2
There are the points of c,such that when they are applied back to the equations,the number of points will become zero.

The point from which the tangents to the circles x2 + y2 – 8x + 40 = 0,  5x2 + 5y2 – 25 x + 80 = 0, x2 + y2 – 8x + 16y + 160 = 0 are equal in length is
  • a)
  • b)
  • c)
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Hansa Sharma answered
The Required point is the radical centre of the three given circles. The radical axes of the three circles taken in pairs are 3x - 24 = 0,16y + 120 = 0 and - 3x + 16y + 80 = 0. On solving, the required point is (8, -15/2).

The locus of the centre of a circle, which touches externally the circle x2 + y2 − 6x − 6y + 14 = 0 and also touches Y-axis, is given by the equation
  • a)
    x2−6x−10y+14=0
  • b)
    x2−10x−6y+14=0
  • c)
    y2−6x−10y+14=0
  • d)
    y2−10x−6y+14=0
Correct answer is option 'D'. Can you explain this answer?

Nipun Tuteja answered
Let the equation of circle be x2+y2+2gx+2fy+c=0.
Then the radius, 
Centre: (−g,−f).
But since circle touches Y-axis, the absolute value of abscissa is equal to the radius, |−g|=r.
⇒g2=r2⇒g2=g2+f2−c
⇒c=f2
Centre of the circle, x2+y2−6x−6y+14=0: (3,3)

As the two circles touch externally, the distance between
centres is equal to the sum of radii. 

Hence, locus of the centre (−g,−f) obtained by substituting −g=x, −f=y⇒g=−x, f=−y, is y2−10x−6y+14=0.

The equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes is:
  • a)
    x2 + y2 - 2ax - 2by = 0
  • b)
    x2 + y2 - ax - by = a2 + b2
  • c)
    x2 + y2 - ax - by = 0
  • d)
    x2 + y2 + 4x + 6y + 13 = 0
Correct answer is option 'C'. Can you explain this answer?

Rithika Sarkar answered
Given information:
- The circle passes through the point (0, 0).
- The circle intercepts on the coordinate axes are a and b.

To find the equation of the circle, we can use the standard form of the equation of a circle:

(x - h)² + (y - k)² = r²

where (h, k) is the center of the circle and r is the radius.

Let's find the center of the circle first:

Since the circle passes through the point (0, 0), the center must lie on the perpendicular bisectors of the intercepts on the coordinate axes.

The perpendicular bisector of the x-axis intercept (a, 0) is the line y = 0.

The perpendicular bisector of the y-axis intercept (0, b) is the line x = 0.

Therefore, the center of the circle is the intersection of these two lines, which is the point (0, 0).

Now, let's find the radius of the circle:

The radius can be found using the distance formula between the center (0, 0) and any point on the circle. Let's take the x-axis intercept (a, 0).

The distance between (0, 0) and (a, 0) is given by:

r = √[(a - 0)² + (0 - 0)²]
= √[a²]
= a

So, the radius of the circle is a.

Therefore, the equation of the circle is:

(x - 0)² + (y - 0)² = a²
x² + y² = a²

However, none of the given options match this equation.

Let's check the options again:

a) x² + y² - 2ax - 2by = 0
b) x² + y² - ax - by = a² + b²
c) x² + y² - ax - by = 0
d) x² + y² + 4x + 6y + 13 = 0

Option C (x² + y² - ax - by = 0) is the closest match to the equation we derived (x² + y² = a²).

Therefore, the correct answer is option C.

The line y = c is a tangent to the parabola 7/2 if c is equal to
  • a)
    a
  • b)
    0
  • c)
    2a
  • d)
    none of these
Correct answer is option 'D'. Can you explain this answer?

Sravya Nair answered
y = x is tangent to the parabola
y=ax2+c
if a= then c=?
y′ =2ax
y’ = 2(7/2)x  =1
x = 1/7
1/7 = 2(1/7)2 + c
c = 1/7 * 2/49
c = 7/2

If the circle x2 + y2 = 9 touches the circle x2 + y2 + 6y + c = 0, then c is equal to
  • a)
    27
  • b)
    36
  • c)
    –36
  • d)
    –27
Correct answer is option 'A'. Can you explain this answer?

Jithin Roy answered

touches the another circle

Now, Central first circle will be
And its radius will be 3 units.
Also,centre of second circle
And radius,

As both touches each other
So,

The equation of the circle passing through (3, 6) and whose centre is (2, –1) is
  • a)
    x2 + y2 – 4x + 2y = 45
  • b)
    x2 + y2 – 4x – 2y + 45 = 0
  • c)
    x2 + y2 + 4x – 2y = 45 
  • d)
    x2 + y2 – 4x + 2y + 45 = 0
Correct answer is option 'A'. Can you explain this answer?

Meghana Pillai answered
To find the equation of the circle passing through (3, 6) and with center (2, c), we can use the distance formula.

The distance between the center (2, c) and the point (3, 6) should be equal to the radius of the circle. Let's call the radius r.

Using the distance formula:
√((3 - 2)^2 + (6 - c)^2) = r

Simplifying this equation:
√(1 + (6 - c)^2) = r

Squaring both sides of the equation:
1 + (6 - c)^2 = r^2

Now we need one more piece of information to find the value of c and r.

The angle between the tangents drawn from the origin to the circle = (x−7)2+(y+1)2 = 25 is 
  • a)
    π/8
  • b)
    π/2
  • c)
    π/6
  • d)
    π/3
Correct answer is option 'B'. Can you explain this answer?

Shalini Yadav answered
Let the equation of tangent drawn from (0,0) to the circle be y=mx. Then, p = a ⇒ 7m+1/(m2+1)1/2= 5
⇒24m2 + 14m−24=0
⇒12m2 + 7m−12=0
⇒m1m2 = −12/12 =−1
∴ Required angle = π/2

Line 3x + 4y = 25 touches the circle x2 + y2 = 25 at the point
  • a)
    (4, 3)
  • b)
    (3, 4)
  • c)
    (–3, –4)
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Anagha Sharma answered
Method to Solve :

Y= 4; x=3

Get X in terms of y from first eq.

X=(25-4y)/3

Substitute in second eq.

By solving you will get

Y2-8y+16=0

Here you will get y value as 4

Substitute in first eq we will get X value as 3

The circles x2+y2+6x+6y = 0 and x2+y2−12x−12y = 0
  • a)
    touch each other externally
  • b)
    touch each other internally
  • c)
    cut orthogonally
  • d)
    intersect in two points
Correct answer is option 'A'. Can you explain this answer?

Gaurav Kumar answered
Given equation of circles are
x2+y2+6x+6y=0....(i)
and x2+y2−12x−12y=0....(ii)
Here, g1 = 3,f2 = 3, g2 = −6 and f2 = −6
∴ Centres of circles are C1(−3,−3) and C2(6,6) respectively and radii are r1 = 3√2 and r2 = 6√2 respectively.
Now, C1C2 = √[(6+3)2 + (6+3)2]
= 9√2
and r1 + r2
​= 3(2)1/2 + 6(2)1/2
= 9(2)1/2
​⇒ C1C2 = r1 + r2
∴ Both circles touch each other externally.

The eccentricity of the conic 9x2 − 16y2 = 144 is
  • a)
    4/3
  • b)
    5/4
  • c)
    √7
  • d)
    4/5
Correct answer is option 'B'. Can you explain this answer?

Mihir Chaudhary answered
+16y2=144 is:

To find the eccentricity of a conic, we need to first identify the type of conic. We can rewrite the given equation as:

9x^2/144 + 16y^2/144 = 1

Dividing both sides by 144, we get:

x^2/16 + y^2/9 = 1

This is the equation of an ellipse. To find the eccentricity, we need to first find the distance between the center of the ellipse and one of its foci. The center of the ellipse is at the point (0,0). We can find the length of the semi-major axis (a) and the semi-minor axis (b) using the equation:

a^2 = 16, b^2 = 9

a = 4, b = 3

The distance between the center and one of the foci (c) can be found using the equation:

c^2 = a^2 - b^2

c^2 = 16 - 9 = 7

c = sqrt(7)

The eccentricity (e) of the ellipse is given by the equation:

e = c/a

e = sqrt(7)/4

Therefore, the eccentricity of the conic 9x^2+16y^2=144 is sqrt(7)/4.

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