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Standard XII

Mathematics

Direct Common Tangent

two circles a...

Question

two circles are externally tangent. Lines PAB and PA'B' are common tangents with A and A' on the smaller circle and B and B' on the larger circle. If PA = AB = 4, then the square of radius of circle is

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Solution

$\frac{A C_{1}}{P A} = \frac{B C_{2}}{P B} \Rightarrow B C_{2} = 2 A C_{1}$
$P C_{1} = \sqrt{16 + r^{2}}$
and $P C_{2} = \sqrt{64 + 4 r^{2}} = 2 \sqrt{16 + r^{2}}$
$P C_{2} - P C_{1} = 3 r$
$\Rightarrow 2 \sqrt{16 + r^{2}} - \sqrt{16 + r^{2}} = 3 r$
$\Rightarrow \sqrt{16 + r^{2}} = 3 r$
$\Rightarrow 16 + r^{2} = 9 r^{2}$
$\Rightarrow r^{2} = 2$

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two circles are externally tangent. Lines PAB and PA'B' are common tangents with A and A' on the smaller circle and B and B' on the larger circle. If PA = AB = 4, then the square of radius of circle is

AC1PA=BC2PB⇒BC2=2AC1PC1=√16+r2and PC2=√64+4r2=2√16+r2PC2−PC1=3r⇒2√16+r2−√16+r2=3r⇒√16+r2=3r⇒16+r2=9r2⇒r2=2