If a circle C passing through (4,0) touches the circle x^2+y^2+4x-6y-12=0 externally at th…

Question

If a circle $C$ passing through $(4,0)$ touches the circle $x^2+y^2+4x-6y-12=0$ externally at the point $(1,-1)$, then the radius of the circle $C$ is:

Accepted Answer

Given circle: centre $A(-2,3)$, radius $=\sqrt{4+9+12}=5$. Since $AP=\sqrt{(1+2)^2+(-1-3)^2}=\sqrt{9+16}=5=r_A$, the point $(1,-1)$ indeed lies on it.

For external tangency, $C$'s centre $B$ lies on ray $AP$ beyond $P$, at distance $r_A+r_C$ from $A$, i.e. distance $r_C$ from $P$ in the same direction $\left(\dfrac35,-\dfrac45\right)$:
$$B=\left(1+\frac{3r_C}{5},\,-1-\frac{4r_C}{5}\right).$$
Since $C$ also passes through $(4,0)$, $|B-(4,0)|=r_C$. Substituting and simplifying:
$$10-2r_C+r_C^2=r_C^2\ \Rightarrow\ r_C=5.$$

Solutions · 1