Question

If the circles x² + y² + 2ax + c = 0 and x² + y² + 2by + c = 0 touch each other, then
(a) $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}$

(b) $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$

(c) a + b = 2c

(d) $\frac{1}{a}+\frac{1}{b}=\frac{2}{c}$

Answer 1

(a) $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}$

Given:
x² + y² + 2ax + c = 0 ...(1)
And, x² + y² + 2by + c = 0 ...(2)
For circle (1), we have:
Centre = $\left(-a,0\right)$ = C₁

For circle (2), we have:
Centre = $\left(0,-b\right)$ = C₂

Let the circles intersect at point P.
∴ Coordinates of P = Mid point of C₁C₂
⇒ Coordinates of P = $\left(\frac{-a+0}{2},\frac{0-b}{2}\right)=\left(\frac{-a}{2},\frac{-b}{2}\right)$

Now, we have:
$$\begin{gathered} P{C}_1=\mathrm{radius}\mathrm{of}\left(1\right) \\ \Rightarrow \sqrt{{\left(-a+\frac{a}{2}\right)}^2+{\left(0-\frac{b}{2}\right)}^2}=\sqrt{a^2-c} \\ \Rightarrow {\frac{a}{4}}^2+{\frac{b}{4}}^2=a^2-c...\left(3\right) \end{gathered}$$


$$\begin{gathered} \mathrm{Also},\mathrm{radius}\mathrm{of}\mathrm{circle}\left(1\right)=\mathrm{radius}\mathrm{of}\mathrm{circle}\left(2\right) \\ \Rightarrow \sqrt{a^2-c}=\sqrt{b^2-c} \\ \Rightarrow a^2=b^2...\left(4\right) \end{gathered}$$

From (3) and (4), we have:

$$\begin{gathered} \frac{a^2}{2}=a^2-c \\ \Rightarrow \frac{a^2}{2}=c \\ \Rightarrow \frac{2}{a^2}=\frac{1}{c} \\ \Rightarrow \frac{1}{a^2}+\frac{1}{a^2}=\frac{1}{c} \\ \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c} \end{gathered}$$