Question

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

Answer 1

Let center of circle is $(h,k)$
So equation of circle
${(x-h)}^2+{(y-k)}^2=r^2$
So, $(3,-2)$ & $(-2,0)$ satisfies the above equation
${(3-h)}^2+{(-2-k)}^2=r^2\to i$
${(-2-k)}^2+{(-k)}^2=r^2\to ii$
As, $r^2=r^2$ so, from $i$ and $ii$
${(3-h)}^2+{(2+k)}^2={(2+h)}^2+k^2$
$9-6h+h^2+4+4k+k^2-4+4h+h^2+k^2$
$=9+4k=10h\to iii$
As, $(h,k)$ lies in $2x-y=3$
$2h-k=3\to iv$
$k=2h-3$

Solving $iii$ and $iv$
$9+4(2h-3)=10h$
$9+8h-12=10h$
$-3=2h$
$h=-\frac{3}{2}$
So, $k=2(-\frac{3}{2})-3$
$=-3-3=-6$
Equation of circle
$={(x+\frac{3}{2})}^2+{(y+6)}^2=r^2$

As (-2,0) lies on the circle, Substituting it.
${(-2+\frac{3}{2})}^2+6^2=r^2$
$r^2=\frac{1}{4}+6^2$
$r=\sqrt{36.25}$

So, ${(x+\frac{3}{2})}^2+{(y+6)}^2=36.25$