Question

Find the equation of the circle which passes through $(3,-2),(-2,0)$ and has its centre on the line $2x-y=3$.

Answer 1

Equation of circle with centre $(h,k)$ is

${(x-h)}^2+{(y-k)}^2=r^2$

Since the circle passes through $(3,-2)$

${(3-h)}^2+{(-2-k)}^2=r^2$

$\Rightarrow 9+h^2-6h+4+k^2+4k=r^2$

$\Rightarrow h^2+k^2-6h+4k+13=r^2$ ........$\left(1\right)$

Since the circle passes through $(-2,0)$

${(-2-h)}^2+{(0-k)}^2=r^2$

$\Rightarrow 4+h^2+4h+k^2=r^2$

$\Rightarrow h^2+k^2+4h+4=r^2$ ........$\left(2\right)$

Eqn$\left(2\right)-$Eqn$\left(1\right)$

$\Rightarrow h^2+k^2+4h+4-h^2-k^2+6h-4k-13=r^2-r^2$

$\Rightarrow 10h-4k=9$ ........$\left(3\right)$

Since center $(h,k)$ lies on the line $2x-y=3$

$2h-k=3$ ............$\left(4\right)$

Eqn$\left(3\right)-$Eqn$\left(4\right)$

$10h-4k-2h+k=9-3$

$\Rightarrow 8h-3k=6$ ...........$\left(5\right)$

Solving eqn$\left(4\right)$ and $\left(5\right)$

$8\times$eqn$\left(4\right)-2\times$eqn$\left(5\right)$

$16h-8k-16h+6k=24-12$

$\Rightarrow -2k=12$ or $k=-6$

From $\left(5\right)$ we have

$8h=6+3k$

$=6-3\times 6=6-18=-12$

$\therefore h=\frac{-12}{8}=\frac{-3}{2}$

Let us find the radius of the circle using eqn$\left(1\right)$

${(3+\frac{3}{2})}^2+{(-2+6)}^2=r^2$

$\Rightarrow r^2=\frac{81}{4}+16=\frac{81+64}{4}=\frac{145}{4}$

$\therefore$ equation of the circle is ${(x+\frac{3}{2})}^2+{(y+6)}^2=\frac{145}{4}$

or ${(2x+3)}^2+4{(y+6)}^2=145$