Question

Find the equations of the circle having the pair of lines $x^2+2xy+3x+6y=0$ its normals and having the size just sufficient to contain the circle $x(x-4)+y(y-3)=0$

Answer 1

$\begin{array}{c}{x}^2+2xy+3x+6y=0\\ x^2+3x+2xy+6y=0\\ x\left(3+x\right)+2y\left(x+3\right)=0\\ \left(x+2y\right)\left(x+3\right)=0\\ x=-3\\ orx=-2y,y=\frac{3}{2}\\ x\left(x-4\right)+y\left(y-3\right)=0\end{array}$

Points on circle is $[4,3]$

then equation of circle

$\begin{array}{c}\left(x+3\right)\left(x-4\right)+\left(y-\frac{3}{2}\right)\left(y-3\right)=1\\ x^2-x-12+y^2-\frac{9y}{2}+\frac{9}{2}=0\\ x^2+y^2-x-\frac{9y}{2}=\frac{15}{2}\end{array}$