Question

Find the equation of the circle whose centre is $(2,-3)$ and which passes through the intersection of the lines $3x+2y=11$ and $2x+3y=4$.

Answer 1

Given equations are:

$3x+2y=11$ ……..(1)

$2x+3y=4$ ………(2)

On solving $\left(1\right)$ & $\left(2\right)$, we get $x=5$ and $y=-2$

Point of intersection are $=(5,-2)$

Therefore, radius of circle is distance between center and point of intersection and center of circle is given i.e $(2,-3)$

$R=\sqrt{10}$

Hence required equation is :

$(x-2{)}^2+(y+3{)}^2=(\sqrt{10}{)}^2$

$x^2+y^2-4x+6y+3=0$

Hence, this is the answer.