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Circle — BITSAT Mathematics PYQs

27 previous year questions from Circle with answers and solutions. Numbered list, year tags, and one-tap solutions — built for serious JEE / NEET practice.

27 questionsMathematicsSolutions on every page
1

The locus of the mid-point of a chord of the circle x^2+y^2=4 , which subtends a right angle at the origin is

BITSAT 2024 Solution
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2

If p and q be the longest and the shortest distance respectively of the point (-7,2) from any point ( , ) on the curve whose equation is x^2+y^2-10 x-14 y-51=0 , then G.M. of p

BITSAT 2024 Solution
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3

From a point A (0,3) on the circle (x+2)^2+(y-3)^2=4 , a chord A B is drawn and it is extended to a point Q such that A Q=2 A B . Then the locus of Q is

BITSAT 2024 Solution
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4

The equation of the circle with centre (0,2) and radius 2 is x^2+y^2-m y=0 . The value of m is

BITSAT 2023 Solution
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5

A circle touches both the y -axis and the line x+y=0 . Then the locus of its center is

BITSAT 2023 Solution
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6

For each parabola y=x^2+p x+q , meeting coordinate axes at 3-distinct points, if circles are drawn through these points, then the family of circles must pass through

BITSAT 2022 Solution
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7

The locus of the mid-point of the chord if contact of tangents drawn from points lying on the straight line 4 x-5 y=20 to the circle x^2+y^2=9 is

BITSAT 2022 Solution
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8

The equation of the circle, which touches the line y=5 and passes through (-1,2) and (1,2) is

BITSAT 2021 Solution
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9

If a chord of the circle x²+y²=8 makes equal intercets of length a on the coordinate axes, then

BITSAT 2020 Solution
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10

In the given figure, the equation of the larger circle is x²+y²+4 y-5=0 and the distance between centres is 4 . Then the equation of smaller circle is

BITSAT 2020 Solution
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11

If the coordinates at one end of a diameter of the circle x²+y²-8 x-4 y+c=0 are (-3,2) , then the coordinates at the other end are

BITSAT 2020 Solution
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12

The area of the triangle formed by joining the origin to the point of intersection of the line x 5 + 2 y = 3 5 and circle x 2 + y 2 = 10 is

BITSAT 2018 Solution
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13

Radius of the largest circle which passes through the focus of the parabola y 2 = 4 x and contained in it, is

BITSAT 2018 Solution
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14

If two distinct chords drawn from the point ( p , q ) on the circle x 2 + y 2 = p x + q y (where p q ≠ 0 ) are bisected by the X -axis, then

BITSAT 2017 Solution
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15

The line joining (5,0) to ((10 , 10 ) is divided internally in the ratio 2: 3 at P . If varies, then the locus of P is

BITSAT 2016 Solution
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16

The number of integral values of for which x ²+ y ²+ x +(1- ) y +5=0 is the equation of a ircle whose radius cannot exceed 5, is

BITSAT 2016 Solution
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17

The lengths of the tangent drawn from any point on the circle 15 x²+15 y²-48 x+64 y=0 to the two circles 5 x²+5 y²-24 x+32 y+75=0 and 5 x²+5 y²-48 x+64 y+300=0 are in the ratio of

BITSAT 2016 Solution
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18

The length of the chord x+y=3 intercepted by the circle x²+y²-2 x-2 y-2=0 is

BITSAT 2016 Solution
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19

Area of the circle in which a chord of length 2 makes an angle / 2 at the centre, is

BITSAT 2015 Solution
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20

The angle of intersection of the two circles x²+y²-2 x-2 y=0 and x²+y²=4, is

BITSAT 2014 Solution
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21

A pair of tangents are drawn from the origin to the circle x²+y²+20(x+y)+20=0, then the equation of the pair of tangent are

BITSAT 2013 Solution
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22

The length of the tangent drawn from any point on the circle x²+y²+2 f y+ =0 to the circle x ²+ y ²+2 fy + =0, where > >0, is

BITSAT 2012 Solution
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23

If the line x +y =p represents the common chord of the circles x²+y²=a² and x ²+ y ²+ b ²( a > b ), where A and B lie ont he first circle and P and Q lie on the second circle, then

BITSAT 2012 Solution
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24

For what value of (k ) the circles (x^2+y^2+5 x+3 y+7=0 ) and (x^2+y^2-8 x+6 y+k=0 ) cuts orthogonally

BITSAT 2011 Solution
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25

If the lines (3 x-4 y+4=0 ) and (6 x-8 y-7=0 ) are tangents to a circle, then the radius of the circle is

BITSAT 2011 Solution
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26

The line (a x+b y=1 ) cuts ellipse (c x^2+d y^2=1 ) only once if

BITSAT 2010 Solution
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27

If (|r| > ) land (x=a+ a r + a r^2 + ) to ( ), ( y = b - b r + b r ^2 - . to ) and (z=c+ c r^2 + c r^4 + ) to ( ), then ( x y z = )

BITSAT 2009 Solution
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