Question

If the equations of the circles whose radii are $r$ and $R$ be respectively $S=0$ and $S^{\prime }=0$, then prove that the circles $\frac{S}{r}\pm \frac{S^{\prime }}{R}=0$ will cut orthogonally.

Answer 1

For the sake of convenience let the line of centres be chosen as axis of $x$ and distance between them be $2a$ and mid-point of the centres be chosen as origin so that the centres are $(a,0)$, radius $r$ for $S$ and $(-a,0)$, radius $R$ for $S^{\prime }$.
$\therefore S=(x-a{)}^2+y^2=r^2$
or $x^2+y^2-2ax+a^2-r^2=0$
$S^{\prime }=(x+a{)}^2+y^2=R^2$
or $x^2+y^2+2ax+a^2-R^2=0$.
The two circles are
$\frac{S}{r}\pm \frac{S^{\prime }}{R}=0$ or $RS\pm r{S}^{\prime }=0$
Circle $RS+r{S}^{\prime }=0$ becomes
$(R+r)(x^2+y^2+a^2)-2ax(R-r)-rR(r+R)=0$
or $x^2+y^2-2a\frac{R-r}{R+r}x+(a^2-rR)=0.....\left(1\right)$
Replacing $r$ by $-r$ the circle $RS-r{S}^{\prime }=0$ becomes
$x^2+y^2-2a\frac{R+r}{R-r}x+(a^2+rR)=0.....\left(2\right)$
The circle $\left(1\right)$ and $\left(2\right)$ will cut orthogonally if
$2g_1{g}_2+2f_1{f}_2=c_1+c_2$
i.e. $2\{-a\frac{R-r}{R+r}\}\{-a\frac{R+r}{R-r}\}+0=(a^2-rR)+(a^2+rR)$
or $2a^2=2a^2$ which is true.
Hence the circles $\frac{S}{r}\pm \frac{S^{\prime }}{R}$ cut each other orthogonally.