Question

A circle $x^2+y^2+4x-2\sqrt{2}y+c=0$ is the bisector circle of the circles $S_1,$ and $S_1$ is the bisector circle of circles $S_2,$ and so on. If the sum of radii of all these circles is $2$, then the value of $c$ is $k\sqrt{2}$, where the value of $k$ is.............

Answer 1

The given equation of circle $x^2+y^2+4x-2\sqrt{2}y+c=0$ can be written as

$x^2+4x+4+y^2-2\sqrt{2}y+2+c-6=0$

or, $(x+2{)}^2+(y-\sqrt{2}{)}^2=6-c$.

This circle has centre at $(-2,\sqrt{2})$. and radius $\sqrt{6-c}$ units.

Since this circle is the bisector circle of $S_1$ then radius of $S_1$ be $\frac{\sqrt{6-c}}{2}$.

Similarly the radius of $S_2$ be $\frac{\sqrt{6-c}}{2^2}$, and so on.

According to the problem,

$\sqrt{6-c}(1+\frac{1}{2}+\frac{1}{2^2}+.......)=2$

or, $\sqrt{6-c}\times 2=2$

or, $6-c=1$

or, $c=5$

or, $k\sqrt{2}=5$ [Since $c=k\sqrt{2}$]

or, $k=\frac{5}{\sqrt{2}}$.