If a circle passes through the point $(a, b)$ and cuts the circle $x^2+y^2=p^2$ orthogonally, then the…
If a circle passes through the point $(a, b)$ and cuts the circle $x^2+y^2=p^2$ orthogonally, then the equation of the locus of its centre is
$x^2+y^2-3 a x-4 b y+\left(a^2+b^2-p^2\right)=0$
$2 a x+2 b y-\left(a^2-b^2+p^2\right)=0$
$x^2+y^2-2 a x-3 b y+\left(a^2-b^2-p^2\right)=0$
$2 a x+2 b y-\left(a^2+b^2+p^2\right)=0$
Solution
Let the centre be $(\alpha, \beta)$
$\because$ It cut the circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{p}^2$ orthogonally
$2(-\alpha) \times 0+2(-\beta) \times 0=c_1-p^2$
$\mathrm{c}_1=\mathrm{p}^2$
Let equation of circle is $x^2+y^2-2 \alpha x-2 \beta y+p^2=0$
It pass through $(a, b) \Rightarrow a^2+b^2-2 \alpha a-2 \beta b+p^2=0$
Locus $\therefore 2 a x+2 b y-\left(a^2+b^2+p^2\right)=0$