Circle And System Of Circles Question 191
Question: The length of common chord of the circles $ {{(x-a)}^{2}}+y^{2}=a^{2} $ and $ x^{2}+{{(y-b)}^{2}}=b^{2} $ is
[MP PET 1989]
Options:
A) $ 2\sqrt{a^{2}+b^{2}} $
B) $ \frac{ab}{\sqrt{a^{2}+b^{2}}} $
C) $ \frac{2ab}{\sqrt{a^{2}+b^{2}}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Equation of common chord is $ ax-by=0 $ . Now length of common chord $ =2\sqrt{r_1^{2}-p_1^{2}}=2\sqrt{r_2^{2}-p_2^{2}} $ where $ r_1 $ and $ r_2 $ are radii of given circles and $ p_1,\ p_2 $ are the perpendicular distances from centres of circles to common chords.
Hence required length $ =2\sqrt{a^{2}-\frac{a^{4}}{a^{2}+b^{2}}}=\frac{2ab}{\sqrt{a^{2}+b^{2}}} $ .