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 to

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