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JEE MATHEMATICS MCQ: CIRCLES

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Application +4 | -1
If circles x2+y2+2ax+c2=0x^2+y^2+2 a x+c^2=0 and x2+y2+2by+c2=0x^2+y^2+2 b y+c^2=0, touch each other then
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Correct option: B
1a2+1b2=1c2\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}
Option B (1a2+1b2=1c2\frac{{1}}{{a^2}}+\frac{{1}}{{b^2}}=\frac{{1}}{{c^2}}) is correct because it is derived from the condition that the distance between the centers equals the sum or difference of the radii for touching circles.

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