Question

Find the equation of the circle passing through the points $(3,4),(3,-6)$ and $(-1,2)$.

Answer 1

Let centre of circle be $(\alpha ,\beta )$

$\therefore (\alpha -3{)}^2+(\beta -4{)}^2=(\alpha -3{)}^2+(\beta +6{)}^2$

$\Rightarrow {\beta }^2-8\beta +16={\beta }^2+12\beta +36$

$\Rightarrow 20\beta =-20$

$\Rightarrow \beta =-1$

And

$(\alpha -3{)}^2+(-1-4{)}^2=(\alpha +1{)}^2+(-1-2{)}^2$

$\Rightarrow {\alpha }^2-6\alpha +9+25={\alpha }^2+2\alpha +1+9$

$\Rightarrow 8\alpha =24$

$\Rightarrow \alpha =3$

$\therefore$ equation of circle : $(x-3{)}^2+(y+1{)}^2=(3-3{)}^2+(4+1{)}^2$

$\Rightarrow (x-3{)}^2+(y+1{)}^2=25$