$P Q$ is a tangent to circles with centres $A$ and $B$ at $P$ and $Q$ respectively. If $A B = 10 c m$ and $P Q = 8 c m$ , and area of triangle $A P O$ is four times the area of triangle $O Q B$ , the radius of the bigger circle is

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Solution

In, $△ A P O$ and $△ B Q O$

$∠ A P O = ∠ B Q O = 90^{\circ}$ (radius is perpendicular to tangent)

$∠ A O P = ∠ B O Q$ (vertically opposite angles)

$△ A P O \sim △ B Q O$ (by A.A axiom)

$⟹ \frac{A r . (△ A P O)}{A r . (△ B Q O)} = \frac{A P^{2}}{B Q^{2}}$ (Areas related to similar triangles)

$⟹ \frac{4 A r . (△ B Q O)}{A r . (△ B Q O)} = (\frac{A P}{B Q})^{2}$

$⟹ \frac{A P}{B Q} = 2$

$⟹ A P = 2 B Q$

$⟹ r_{1} = 2 r_{2}$

Length of traversal common tangent $P Q = \sqrt{A B^{2} - (A P + B Q)^{2}}$

$⟹ 8 = \sqrt{10^{2} - (r_{1} + r_{2})^{2}}$

$⟹ 64 = 100 - (r_{1} + r_{2})^{2}$

$⟹ (r_{1} + r_{2}) = 100 - 64 = 36 = 6^{2}$

$⟹ r_{1} + r_{2} = 6$

$⟹ 2 r_{2} + r_{2} = 6$ ( $∵ r_{1} = 2 r_{2}$ )

$⟹ 3 r_{2} = 6$

$⟹ r_{2} = \frac{6}{3} = 2 c m$

$∴ r_{1} = 2 r_{2} = 2 \times 2 = 4 c m$

The radius of bigger circle $= 4 c m$

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PQ is a tangent to circles with centres A and B at P and Q respectively. If AB=10 cm and PQ=8 cm, and area of triangle APO is four times the area of triangle OQB, the radius of the bigger circle is

$P Q$ is a tangent to circles with centres $A$ and $B$ at $P$ and $Q$ respectively. If $A B = 10 c m$ and $P Q = 8 c m$ , and area of triangle $A P O$ is four times the area of triangle $O Q B$ , the radius of the bigger circle is