Question:
Three circles of radii a, b, c \((a \lt b \lt c)\) touch each other externally. If they have x-axis as a common tangent, then :
Three circles of radii a, b, c \((a \lt b \lt c)\) touch each other externally. If they have x-axis as a common tangent, then :
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Two circles of radii \(r_1,r_2\) touching externally and both touching the x-axis have their centres \(2\sqrt{r_1r_2}\) apart horizontally. The smallest circle sits between the other two.
Updated On: Nov 5, 2026
- \(a, b, c\) are in A.P.
- \(\sqrt{a},\sqrt{b},\sqrt{c}\) are in A.P
- \(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}\)
- \(\frac{1}{\sqrt{b}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}\)
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The Correct Option is C
Solution and Explanation
Two externally tangent circles of radii \(r_1,r_2\) touching the x-axis: distance between centres \(=r_1+r_2\), vertical gap \(=|r_1-r_2|\), so horizontal gap \(=\sqrt{(r_1+r_2)^2-(r_1-r_2)^2}=2\sqrt{r_1r_2}\).
Three mutually touching circles on the same side of the x-axis: the smallest (radius \(a\)) lies in the gap between the other two (radii \(b\) and \(c\)).
Hence horizontal distance between the \(b\)- and \(c\)-circles \(=\) sum of the other two gaps: \(2\sqrt{bc}=2\sqrt{ab}+2\sqrt{ac}\).
Divide by \(2\sqrt{abc}\): \(\frac{1}{\sqrt{a}}=\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{b}}\).
Answer: (3) \(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}\)
Three mutually touching circles on the same side of the x-axis: the smallest (radius \(a\)) lies in the gap between the other two (radii \(b\) and \(c\)).
Hence horizontal distance between the \(b\)- and \(c\)-circles \(=\) sum of the other two gaps: \(2\sqrt{bc}=2\sqrt{ab}+2\sqrt{ac}\).
Divide by \(2\sqrt{abc}\): \(\frac{1}{\sqrt{a}}=\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{b}}\).
Answer: (3) \(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}\)
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