Question

Question 54
Two circles of radii 15 cm and 12 cm intersect each other, and the length of their common chord is 18
cm. What is the distance (in cm) between their centres ?
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Answer 1

Two circles of radii 15 cm and 12 cm intersect each other, and the length of their common chord is 18

cm. What is the distance (in cm) between their centres ?

Here , Radius (OA) = 15 cm, and

Radius (O'A) = 12 cm ;

Length of common chord (AB) = 18 cm

Join two centres which intersects AB at M .

$AM = MB = \frac{AB}{2} = \frac{18}{2} = 9\:cm$

$i) \triangle AMO \:is \: right \:triangle . \\AMO = 90\degree$

$OA = 15 \:cm, AM = 9 \:cm$

$OA^{2} = OM^{2} + MA^{2}$

$\blue { ( By \: Pythagoras \:Theorem ) }$

$\implies 15^{2} = OM^{2} - 9^{2}$

$\implies OM^{2} = 225 - 81 = 144$

$\implies OM = 12 \:cm$

$ii) \triangle AMO' \:is \: right \:triangle . \\AMO' = 90\degree$

$O'A = 12 \:cm, AM = 9 \:cm$

$O'A^{2} = O'M^{2} + MA^{2}$

$\blue { ( By \: Pythagoras \:Theorem ) }$

$\implies 12^{2} = OM^{2} - 9^{2}$

$\implies O'M^{2} = 144 - 81 = 63$

$\implies O'M = 3\sqrt{7} \:cm$

$Now, Distance \: between \:the \: centers \\= OO' \\= OM + MO' \\= 12 \:cm + 3 \sqrt{7} \:cm$

Therefore.,

$\red { Distance \: between \:the \: centres }\\\green { 12 \:cm + 3 \sqrt{7} \:cm }$

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