Circle is an important topic in the Mathematics section in BITSAT exam. Practising this topic will increase your score overall and make your conceptual grip on BITSAT exam stronger.
This article gives you a full set of BITSAT PYQs for Circle with explanations for effective preparation. Practice of BITSAT Mathematics PYQs including Circle questions regularly will improve accuracy, speed, and confidence in the BITSAT 2026 exam.
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BITSAT PYQs for Circle with Solutions
1.
From a point A(0,3) on the circle \[ (x + 2)^2 + (y - 3)^2 = 4 \] a chord AB is drawn and extended to a point Q such that AQ = 2AB. Then the locus of Q is:- \( (x + 4)^2 + (y - 3)^2 = 16 \)
- \( (x + 1)^2 + (y - 3)^2 = 32 \)
- \( (x + 1)^2 + (y - 3)^2 = 4 \)
- \( (x + 1)^2 + (y - 3)^2 = 1 \)
2.
The radius of the circle with the polar equation $r^2 - 8r( \sqrt{3} \, \cos \, \theta + \sin \, \theta) + 15 = 0$ is- 8
- 7
- 6
- 5
3.
If the straight line $2x + 3y - 1 = 0$, $x + 2y - 1 = 0$ and $ax + by - 1 = 0$ form a triangle with origin as orthocentre, then $(a,b)$ is equal to:- (6,4)
- (-3,3)
- (-8,8)
- (0,7)
4.
If \( p \) and \( q \) be the longest and the shortest distance respectively of the point (-7,2) from any point (\(\alpha, \beta\)) on the curve whose equation is \[ x^2 + y^2 - 10x - 14y - 51 = 0 \] then the geometric mean (G.M.) of \( p \) is:- \( 2\sqrt{11} \)
- \( 5\sqrt{5} \)
- 13
- 11
5.
A(3,2,0), B(5,3,2), C(-9,6,-3) are three points forming a triangle. AD, the bisector of angle $BAC$ meets BC in D. Find the coordinates of D:- $\left( \frac{19}{8}, \frac{57}{15}, \frac{57}{15} \right)$
- $\left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right)$
- (2,3,0)
- (4,5,6)
6.
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:- $x + y = 2$
- $x^2 + y^2 = 1$
- $x^2 + y^2 = 2$
- $x + y = 1$
7.
The locus of centre of a circle which passes through the origin and cuts off a length of $4$ unit from the line $x = 3$ is- $y^2 + 6x = 0$
- $y^2 + 6x = 13$
- $y^2 + 6x = 10$
- $x^2 + 6y = 13$
8.
If the lines $2x - 3y = 5$ and $3x - 4y = 7$ are two diameters of a circle of radius $7$, then the equation of the circle is- $ x^2 + y^2 + 2x - 4y - 47 = 0$
- $x^2 + y^2 = 49$
- $x^2 + y^2 -2x + 2y -47 = 0$
- $x^2 + y^2 = 17$
9.
The distance from the origin to the image of $(1,1)$ with respect to the line $x + y + 5 = 0$ is:- $7\sqrt{2}$
- $3\sqrt{2}$
- $6\sqrt{2}$
- $4\sqrt{2}$
10.
The maximum area of a rectangle inscribed in a circle of diameter \( R \) is:- \( R^2 \)
- \( \frac{R^2}{2} \)
- \( \frac{R^2}{4} \)
- \( \frac{R^2}{8} \)
11.
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- $x + y = 2$
- $x^{2}+y^{2}=1$
- $x^{2}+y^{2}=2$
- $x + y = 1$
12.
The locus of the point of intersection of the lines \(x = a(1 - t^2)/(1 + t^2)\) and \(y = 2at/(1 + t^2)\) (t being a parameter) represents:- Circle
- Parabola
- Ellipse
- Hyperbola
13.
The inverse of the point $(1, 2)$ with respect to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$, is- $1 , \frac{1}{2}$
- (2, 1)
- (0, 1)
- (1, 0)
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