SSC CPO SI and ASI 2014 Paper

Question : 138 of 200

Marks: +1, -0

Two circles with centres A and B of radii 5 cm and 3 cm, respectively, touch each other internally. If the perpendicular bisector of AB meets the bigger circle at P and Q, then the value of PQ is

$√6$ cm
$2 √6$ cm
$3 √6$ cm
$4 √6$ cm

Validate

Solution:

(4)From

PA = 5cm and AE = 1cm. Now,

{PE} =√{\ {PA}^{2}- {AE}^{2}}=√{5^{2}-1^{2}}

=√{24}=2 √{6} {cm}

Since, perpendicular bisector from centre to the chord bisects it, hence, PQ = 2PE =

4 √6

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