Question

Find the equation of the circle passing through the points $(2,3)$ and $(-1,1)$ an whose center is on the line $x-3y-11=0$ .

Answer 1

Let the equation of the circle

${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2.............\left(A\right)$

Since the circle passes through points $(2,3)$ points $(2,3)$ will satisfy the equation of a circle

Putting $x=2,y=3$ in $\left(A\right)$

$\begin{array}{c}{\left(2-h\right)}^2+{\left(3-k\right)}^2=r^2\\ {\left(2\right)}^2+{\left(h\right)}^2-2\left(2\right)\left(h\right)+{\left(3\right)}^2+k^2-2\left(3\right)\left(k\right)=r^2\\ 4+h^2-4h+9+k^2-6k=r^2\\ h^2+k^2-4h-6k+4+9=r^2\\ h^2+k^2-4h-6k+13=r^2......\left(1\right)\end{array}$

Point $(-1,1)$ will satisfy the equation of circle

Putting $x=-1,y=-1$ in $\left(A\right)$

$\begin{array}{c}{\left(-1-h\right)}^2+{\left(1-k\right)}^2=r^2\\ {\left(-1\right)}^2+{\left(1+h\right)}^2+{\left(1-k\right)}^2=r^2\\ {\left(1+h\right)}^2+{\left(1-k\right)}^2=r^2\\ {\left(1\right)}^2+h^2+2h+1+k^2-2k=r^2\\ h^2+k^2+2h-2k+1+1=r^2\\ h^2+k^2+2h-2k+2=r^2......\left(2\right)\end{array}$

Since center $(h,k)$ lie on the circle $x-3y-11=0$

Point $(h,k)$ will satisfy the equation of line $x-3y-11=0$

So, $h-3k-11=0$

$h-3k=11$