# In Fig. 10.93, there are two concentric circles with centre O. PR and PQS are tangents to the inner circle from point plying on the outer circle. If PR = 7.5 cm, then PS is equal toA. 10 cmB. 12 cmC. 15 cmD. 18 cm

Given:

PR = 7.5 cm

Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By property 1, ∆OSQ is right-angled at OQS (i.e., OQS = 90°) and OPQ is right-angled at OQP (i.e., OQP = 90°).

OQ PS

PO = OS [radius of circle]

POS is an isosceles triangle

Now,

POS is an isosceles triangle and OQ is perpendicular to its base

OQ bisects PS

i.e., PQ = QS

By property 2,

PR = PQ = 7.5 cm (tangent from P)

Now,

PS = PQ + QS

PS = PQ + PQ [ PQ = QS]

PS = 7.5 cm + 7.5 cm

PS = 15 cm

Hence, PS = 15 cm

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