Test: Conic Sections - 1 - JEE MCQ

# Test: Conic Sections - 1 - JEE MCQ

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## 25 Questions MCQ Test - Test: Conic Sections - 1

Test: Conic Sections - 1 for JEE 2024 is part of JEE preparation. The Test: Conic Sections - 1 questions and answers have been prepared according to the JEE exam syllabus.The Test: Conic Sections - 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Conic Sections - 1 below.
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Test: Conic Sections - 1 - Question 1

### If the line 2x – y + λ = 0 is a diameter of the circle x2+y2+6x−6y+5 = 0 then λ =

Detailed Solution for Test: Conic Sections - 1 - Question 1

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

Test: Conic Sections - 1 - Question 2

### Length of common chord of the circles x2+y2+2x+6y = 0 and x2+y2−4x−2y−6 = 0 is

Detailed Solution for Test: Conic Sections - 1 - Question 2

Given equation of circles are

S1 = x2 + y2 + 2x + 6y = 0 ..................1

S2 = x2 + y2 − 4x − 2y − 6 = 0 ............2

Subtract equation 1 - equation 2, we get

S1 - S2 = 6x + 8y + 6 = 0

=> 6x + 8y + 6 = 0 ............3

Cente of the circle S2 is (2, 1)

Now, the length of the perpendicular from the center (2, 1) of of the circle 2 upon the common chord 3 is

l = (6*2 + 8*1 + 6)/√{62 + 82 }

=> l = (12 + 8 + 6)/√{36 + 64}

=> l = 26/√100

=> l = 26/10

=> l = 13/5

Radius of the circle 2 is

r = √{22 + 12 - (-6)}

=> r = √{4 + 1 + 6}

=> r = √11

Now length of common chord = 2√{r2 - l2 }

= 2√{(√11)2 - (13/5)2 }

= 2√{11 - 169/25}

= 2√{(275 - 169)/25}

= 2√{106/25} unit

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Test: Conic Sections - 1 - Question 3

### The circles x2+y2+6x+6y = 0 and x2+y2−12x−12y = 0

Detailed Solution for Test: Conic Sections - 1 - Question 3

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.

Test: Conic Sections - 1 - Question 4

The equation x2+y2 = 0 represents

Test: Conic Sections - 1 - Question 5

The equation 3 (x2+y2)+5x−7y−2 = 0 represents

Test: Conic Sections - 1 - Question 6

Circumcentre of the triangle, whose vertices are (0, 0), (6, 0) and (0, 4) is

Test: Conic Sections - 1 - Question 7

If (x2−a)2+(y−b)2 = c2 represents a circle, then

Detailed Solution for Test: Conic Sections - 1 - Question 7

The standard equation of a circle is given by:

(x-a)2 + (y-b)2 = c2

Where (a,b) is the coordinates of center of the circle and c is the radius.
if c become zero then the radius of circle become zero then there will be no circle exist hence the radius can not be zero. So the correct option is D

Test: Conic Sections - 1 - Question 8

The length of the chord joining the point (4 cos θ, 4 sin θ) and 4 (cos(θ+60o), 4 sin(θ + 60o)) of the circle x2+y2 = 16 is

Test: Conic Sections - 1 - Question 9

Two perpendicular tangents to the circle x2+y2 = r2 meet at P. The locus of P is

Detailed Solution for Test: Conic Sections - 1 - Question 9

Test: Conic Sections - 1 - Question 10

The number of tangents to the circle x2+y2−8x−6y+9 = 0, which pass through the point (3, - 2), is

Test: Conic Sections - 1 - Question 11

The value of k, such that the equation = 2x2+2y2−6x+8y+k = 0 represents a point circle, is

Test: Conic Sections - 1 - Question 12

The equation ax2+by2+2hxy+2gx+2fy+c = 0 represents a circle only if

Detailed Solution for Test: Conic Sections - 1 - Question 12

ax2+by2+2hxy+2gx+2fy+c = 0
In the circle equation,
Coefficients of x2 & y2 are the same and non-zero.
Therefore, a = b ≠ 0
There is no term for xy. Therefore, h = 0.
Radius of the circle: r2 = g2 + f2 – c,
So, g2 + f2 – c > 0

Test: Conic Sections - 1 - Question 13

x2+y2−6x+8y−11 = 0 is a circle. The points (0, 0) and (1, 8) lie

Test: Conic Sections - 1 - Question 14

The length of tangent from the point (2, - 3) to the circle 2x2+2y2 = 1 is

Detailed Solution for Test: Conic Sections - 1 - Question 14

Length of tangent from an external point (h,k) to  S = √(S(h,k))
S(2,-3) = √[2(2)2 + 2(-3)2 - 1]
=> S(2,-3) = √[8 + 18 - 1]
=> S(2,-3) = √25
=> S(2,-3) = 5

Test: Conic Sections - 1 - Question 15

Which one of the following lines is farthest from the centre of the circle x2+y2 = 10?

Test: Conic Sections - 1 - Question 16

Which of the following lines is a normal to the circle (x−1)2+(y−2)2 = 10

Test: Conic Sections - 1 - Question 17

A circle with its centre on the line y = x + 1 is drawn to pass through the origin and touch the line y = x + 2. The centre of the circle is

Test: Conic Sections - 1 - Question 18

Four distinct points (2λ,3λ),(1,0),(0,1) and (0, 0) lie on a circle for

Test: Conic Sections - 1 - Question 19

The locus of the point of intersection of the lines x cos α + y sin α = a and x sin α - y cos α = b is

Test: Conic Sections - 1 - Question 20

The line y = m x + c is a normal to the circle x2+y2+2gx+2fy+c = 0 if

Test: Conic Sections - 1 - Question 21

The line 3x – 4y = 0

Test: Conic Sections - 1 - Question 22

The equations x = a cos θ + b sin θ , and y = a sin θ − b cos θ y = a sin ⁡θ − b cos⁡ θ , 0 ≤ θ ≤ 2 π represent

Test: Conic Sections - 1 - Question 23

The number of points which have the same power w.r.t. two (different) concentric circles is

Test: Conic Sections - 1 - Question 24

A circle passes through (0, 0) ( a, 0), (0, b). The coordinates of its centre are

Test: Conic Sections - 1 - Question 25

The focus of the parabola x2−8x+2y+7 = 0 is

Detailed Solution for Test: Conic Sections - 1 - Question 25

Parabola is x2 – 8x + 2y + 7 = 0
∴   (x – 4)2 = – 2y – 7 + 16
∴   (x – 4)2 = – 2[y – (9/2)]
∴   x2 = – 4ay
⇒ x = x – 4, y = y – (9/2) and 2 = 4a
i.e. a = (1/2)
Its focus is given by x = 0 and y = 0 i.e. x – 4 = 0   and     y – (9/3) = 0
∴    x = 4    and y = (9/2)
∴ focus [4, (9/2)].

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