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