The circle on the chord $x\cos \alpha + y\sin \alpha = p$ of the circle ${x^2} + {y^2} = {a^2}$ as diameter has the equation
${x^2} + {y^2} - {a^2} - 2p(x\cos \alpha + y\sin \alpha - p) = 0$
${x^2} + {y^2} + {a^2} + 2p(x\cos \alpha - y\sin \alpha + p) = 0$
${x^2} + {y^2} - {a^2} + 2p(x\cos \alpha + y\sin \alpha + p) = 0$
${x^2} + {y^2} - {a^2} - 2p(x\cos \alpha - y\sin \alpha - p) = 0$
Let $C_i \equiv x^2 + y^2 = i^2 (i = 1,2,3)$ are three circles. If there are $4i$ points on circumference of circle $C_i$. If no three of all the points on three circles are collinear then number of triangles which can be formed using these points whose circumcentre does not lie on origin, is-
Let $S = 0$ is the locus of centre of a variable circle which intersect the circle $x^2 + y^2 -4x -6y = 0$ orthogonally at $(4, 6)$ . If $P$ is a variable point of $S = 0$ , then least value of $OP$ is (where $O$ is origin)
Let the mirror image of a circle $c_{1}: x^{2}+y^{2}-2 x-$ $6 y+\alpha=0$ in line $y=x+1$ be $c_{2}: 5 x^{2}+5 y^{2}+10 g x$ $+10 f y +38=0$. If $r$ is the radius of circle $c _{2}$, then $\alpha+6 r^{2}$ is equal to$.....$
If a circle $C,$ whose radius is $3,$ touches externally the circle, $x^2 + y^2 + 2x - 4y - 4 = 0$ at the point $(2, 2),$ then the length of the intercept cut by circle $c,$ on the $x-$ axis is equal to
Consider a circle $C_1: x^2+y^2-4 x-2 y=\alpha-5$.Let its mirror image in the line $y=2 x+1$ be another circle $C _2: 5 x ^2+5 y ^2-10 fx -10 gy +36=0$.Let $r$ be the radius of $C _2$. Then $\alpha+ r$ is equal to $......$.