Question

If $(4,3)$ and $(-12,-1)$ are end points of a diameter of a circle, then find the equation of the circle.

Answer 1

The equation of the circle will be,

$\begin{array}{c}\Rightarrow \left(x-4\right)\left(x+12\right)+\left(y+1\right)\left(y-3\right)=0\\ \Rightarrow x^2+8x-48+y^2-2y-3=0\\ \Rightarrow x^2+y^2+8x-2y=51\end{array}$