Circle And System Of Circles Question 288
Question: If circles $ x^{2}+y^{2}+2ax+c=0 $ and $ x^{2}+y^{2}+2by+c=0 $ touch each other, then
[MNR 1987]
Options:
A) $ \frac{1}{a}+\frac{1}{b}=\frac{1}{c} $
B) $ \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}} $
C) $ \frac{1}{a}+\frac{1}{b}=c^{2} $
D) $ \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ C_1(-a,\ 0); $
$ C_2(0,\ -b); $
$ R_1(\sqrt{a^{2}-c}); $
$ R_2(\sqrt{b^{2}-c}) $
$ C_1C_2=\sqrt{a^{2}+b^{2}} $
Since they touch each other, therefore $ \sqrt{a^{2}-c}+\sqrt{b^{2}-c}=\sqrt{a^{2}+b^{2}} $
$ \Rightarrow a^{2}b^{2}-b^{2}c-a^{2}c $ = 0 Multiply by $ \frac{1}{a^{2}b^{2}c},\ $ we get $ \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c} $ .