The locus of centre of a circle passing through $(a, b)$ and cuts orthogonally to circle ${x^2} + {y^2} = {p^2}$, is
$2ax + 2by - ({a^2} + {b^2} + {p^2}) = 0$
$2ax + 2by - ({a^2} - {b^2} + {p^2}) = 0$
${x^2} + {y^2} - 3ax - 4by + ({a^2} + {b^2} - {p^2}) = 0$
${x^2} + {y^2} - 2ax - 3by + ({a^2} - {b^2} - {p^2}) = 0$
In the figure shown, radius of circle $C_1$ be $ r$ and that of $C_2$ be $\frac{r}{2}$ , where $r= \frac {1}{3} PQ,$ then length of $AB$ is (where $P$ and $Q$ being centres of $C_1$ $\&$ $C_2$ respectively)
The equation of the image of the circle ${x^2} + {y^2} + 16x - 24y + 183 = 0$ by the line mirror $4x + 7y + 13 = 0$ is
The circle passing through the intersection of the circles, $x^{2}+y^{2}-6 x=0$ and $x^{2}+y^{2}-4 y=0$ having its centre on the line, $2 x-3 y+12=0$, also passes through the point
The two circles ${x^2} + {y^2} - 2x + 6y + 6 = 0$ and ${x^2} + {y^2} - 5x + 6y + 15 = 0$
A circle passes through the origin and has its centre on $y = x$. If it cuts ${x^2} + {y^2} - 4x - 6y + 10 = 0$ orthogonally, then the equation of the circle is