Question

The centres and radii of two given circles are $A(a,0),b$ and $B(-a,0),c$. Show that the points of contact of their common tangents lie on the circle $x^2+y^2=a^2\pm bc$, it being given that $a>b>c$.

Answer 1

The given circles are
$(x-a{)}^2+y^2=b^2.....(1)$
$(x+a{)}^2+y^2=c^2......(2)$
$C_1{C}_2=2a=a+a>bc$
$\because a>b>c$
Hence the circles are non-intersecting.
Let $(h,k)$ be any point on $\left(1\right)$.
$\therefore (h-a{)}^2+k^2=b^2....(A)$
or $h^2+k^2-2ah+a^2=b^2$
Tangent at any point $(h,k)$ to $\left(1\right)$
$(h-a)x+ky-ah+a^2-b^2=0$
If it is a tangent to $2nd$ circle, then applying the condition of tangency with $2nd$ circle, we get
$\frac{-a(h-a)-ah+a^2-b^2}{\sqrt{\left[\right(h-a{)}^2+k^2]}=b}=c$ by $\left(A\right)$
$-2ah+2a^2-b^2=\pm bc$
Putting for $-2ah$ from (A)
$(b^2-k^2-h^2-a^2)+2a^2-b^2=\pm bc$ by (A)
or $h^2+k^2=a^2\pm bc$
$\therefore$ Locus of $(h,k)$ is $x^2+y^2=a^2\pm bc$.