Circle is an important topic in the Mathematics section in VITEEE exam. Practising this topic will increase your score overall and make your conceptual grip on VITEEE exam stronger.
This article gives you a full set of VITEEE PYQs for Circle with explanations for effective preparation. Practice of VITEEE Mathematics PYQs including Circle questions regularly will improve accuracy, speed, and confidence in the VITEEE 2026 exam.
Also Read
VITEEE PYQs for Circle with Solutions
1.
If $(-3, 2)$ lies on the circle $x^2 + y^2 + 2gx + 2fy + c = 0$, which is concentric with the circle $x^2 + y^2 + 6x + 8y - 5 = 0,$ then c is equal to- 11
- -11
- 24
- 100
2.
The equations of the circle which pass through the origin and makes intercepts of lengths $4$ and $8$ on the $x$ and $y$-axes respectively are- $x^{2}+y^{2}\pm4x\pm8y=0$
- $x^{2}+y^{2}\pm2x\pm4y=0$
- $x^{2}+y^{2}\pm8x\pm16y=0$
- $x^{2}+y^{2}\pm x\pm y=0$
3.
The tangent at $(1, 7)$ to the curve $x^2 = y - 6$ touches the circle $x^2 + y^2 + 16x + 12y + c = 0$ at- (6, 7)
- (-6, 7)
- (6, -7)
- (-6, -7)
4.
The point $(3, -4)$ lies on both the circles $x^{2}+y^{2}-2x+8y+13=0$ and $x^{2}+y^{2}-4x+6y+11=0 $ Then, the angle between the circles is- $60^{\circ}$
- $\tan ^{-1}\left(\frac{1}{2}\right)$
- $\tan ^{-1}\left(\frac{3}{5}\right)$
- $135^{\circ}$
5.
The angle of intersection of the circles $x^{2}+y^{2}-x+y-8=0$ and $x^{2}+y^{2}+2 x+2 y-11=0$ is- $tan^{-1} \left(\frac{19}{9}\right)$
- $tan^{-1} \left(19\right)$
- $tan^{-1} \left(\frac{9}{19}\right)$
- $tan^{-1} \left(9\right)$
Comments