Question

Show that the circles $S\equiv x^2+y^2-2x-4y-20=0$, $S^{\prime }\equiv x^2+y^2+6x+2y-90=0$ touch each other internally. Find their point of contact.

Answer 1

$S\equiv x^2+y^2-2x-4y-20=0$

Center of $S=(1,2)$ and radius $=\sqrt{1+4+20}=5$ unit

$S^{\prime }\equiv x^2+y^2+6x+2y-90=0$

Center of $S^{\prime }=(-3,-1)$ and radius $=\sqrt{3^2+1^2+90}=10$ unit

Distance between the centers of two circles is $\sqrt{(1+3{)}^2+(2+1{)}^2}=5$ unit, which is bigger radius minus the smaller radius.

Thus, the two circles touch internally.

To find their point of contact, we find the line which is obtained by subtracting their equations, which is $6x+2y-90-(-2x-4y-20)=0$

i.e. $8x+6y-70=0$ or $4x+3y-35=0$

A point on the line is given by $(t,\frac{35-4t}{3})$

Its distance from the center of the bigger circle must be 10 unit

$\Rightarrow (t+3{)}^2+{\left(\frac{35-4t+3}{3}\right)}^2=100$

$\Rightarrow 9t^2+81+54t+1444+16t^2-304t=900$

$\Rightarrow 25t^2-250t+625=0$

$\Rightarrow t^2-10t+25=0$

$\Rightarrow t=5$

The point of contact thus becomes $(5,5)$