Question

Find the equation of the circle passing through the point $(-3,2)$ and having its centre at the point of intersection of the lines $x-2y=4$ and $2x+5y+1=0$.

Answer 1

$Inter\sec tionofx-2y-4=0and2x+5y+1=0$

$x-2y-4=0$

$\Rightarrow 2x+5y+1=0$

$\Rightarrow \frac{x}{18}=\frac{y}{-9}=\frac{1}{9}$

$x=2,y=-1$

$circlepas\sin gthroughthepoint\left(-3,2\right)andenterat\left(2,-1\right),$

$radius=\sqrt{5^2+{\left(3\right)}^2}=\sqrt{34}$

$Then,equationofcircle,{\left(x-2\right)}^2+{\left(y+1\right)}^2={\left(\sqrt{34}\right)}^2$

$equationofcircle=x^2+y^2-4x+2y+5=29$

Hence, this is the answer.