Question

Find the equation of circle which passes through the origin, has its centre on the line $x+y=4$, and cuts the circle $x^2+y^2-4x+2y+4=0$ orthogonally.

Answer 1

Equation of circle which passes through the origin is $x^2+y^2+2gx+2fy+c=0$ .....$\left(1\right)$

the centre of the circle $\left(1\right)$ is $(-g,-f)$

if the centre lies on the line $x+y=4$

then $-g-f=4\Rightarrow g+f=-4$ .....$\left(2\right)$

the given equation of the orthogonal circle is $x^2+y^2-4x+2y+4=0$ ....$\left(3\right)$

Comparing the circle $\left(2\right)$ with the general equation of the circle, we get

$g_1=-2;f_1=1$ and $c_1=4$

the circle $\left(1\right)$ is orthogonal to circle $\left(2\right)$

$\therefore 2g{g}_1+2f{f}_1=c+c_1$ if two circles $x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$ and $x^2+y^2+2g_{2}x+2f_{2}y+c_2=0$ are orthogonal then $2g_1{g}_2+2f_1{f}_2=c_1+c_2$

$\Rightarrow 2g(-2)+2f\left(1\right)=0+4$

$\Rightarrow -4g+2f=4$ or $-2g+f=2$ .....$\left(4\right)$

Solving eqn$\left(2\right)$ and eqn$\left(4\right)$ we get

$g+2g=-4-2$ or $3g=-6$ or $g=-2$

$f=-4-g=-4-(-2)=-4+2=-2$ by subustituting for $g=-2$

Thus, the equation of the required circle is $x^2+y^2+2\times -2x+2\times -2y=0$

or $x^2+y^2-4x-4y=0$