Question

Find the equation of circle having the lines $x^2+2xy+3x+6y=0$ as its normal and having size just sufficient to contains the circle. $x(x-4)+y(y-3)=0$.

Answer 1

Pair `normals are $(x+2y)(x+3)=0$

$\therefore$ Normals are $x+2y=0$

Point of intersection of the normal in the centre of the required circle i.e., $C_1(-3,-3/2)$ and centre of given xcircle $C_2(2,3/2)$ and radius

$r^2=\sqrt{4+\frac{9}{2}}=\frac{5}{2}$

Let $r_1$ be radius of required circle

$r_1=|C_1+C_2|+r_2\Rightarrow \sqrt{(-3-2{)}^2+(\frac{3}{2}-\frac{-3}{2}{)}^2}+\frac{5}{2}=\frac{15}{2}$

Hence, equation of required cicrle is

$x^2+y^2+6x-3y-54=0$