Question

Find the equation of a circle which touches the line $x+y=5$ at the point $(-2,7)$ and cuts the circle $x^2+y^2+4x-6y+9=0$ orthogonally.

Answer 1

The required circle by Rule $\left(n_1\right)$ is
$(x+2{)}^2+(y-7{)}^2+\lambda (x+y-5)=0$
or $x^2+y^2+x(4+\lambda )+y(-14+\lambda )+(53+5\lambda )=0$
It cuts orthogonally the circle
$2\frac{4+\lambda }{2}.2+2\frac{(-14+\lambda )}{2}(-3)=9+53-5\lambda$
or $\lambda (2-3+5)=62-42-8$
or $4\lambda =12\Rightarrow \lambda =3$
Hence the required circle is
$x^2+y^2+7x-11y+38=0$.