Theorem 9.2 : The lengths of tangents drawn froman external point to a

1.	Theorem 9.2 : The lengths of tangents drawn froman external point to a circle are equalProof : We are given a circle with centre O, a pointlying outside the circle and two tangents PQ, PR on the circle from P (see Fig. 9.7). Were required to prove that PQ PR.Fig. 9.6
Answer» Let two tangent PT and QT are drawn to circle of centre O as shown in figure. Both the given tangents PT and QT touch to the circle at P and Q respectively. We have to proof : length of PT = length of QT Construction :- draw a line segment ,from centre O to external point T { touching point of two tangents } . Now ∆POT and ∆QOT We know, tangent makes right angle with radius of circle. Here, PO and QO are radii . So, ∠OPT = ∠OQT = 90° Now, it is clear that both the triangles ∆POT and QOT are right angled triangle.nd a common hypotenuse OT of these [ as shown in figure ] Now, come to the concept , ∆POT and ∆QOT ∠OPT = OQT = 90° Common hypotenuse OT And OP = OQ [ OP and OQ are radii]So, R - H - S rule of similarity ∆POT ~ ∆QOT Hence, OP/OQ = PT/QT = OT/OT PT/QT = 1 PT = QT [ hence proved]

Theorem 9.2 : The lengths of tangents drawn froman external point to a circle are equalProof : We are given a circle with centre O, a pointlying outside the circle and two tangents PQ, PR on the circle from P (see Fig. 9.7). Were required to prove that PQ PR.Fig. 9.6

Answer»

Let two tangent PT and QT are drawn to circle of centre O as shown in figure. Both the given tangents PT and QT touch to the circle at P and Q respectively.

We have to proof : length of PT = length of QT Construction :- draw a line segment ,from centre O to external point T { touching point of two tangents } .

Now ∆POT and ∆QOT We know, tangent makes right angle with radius of circle. Here, PO and QO are radii . So, ∠OPT = ∠OQT = 90° Now, it is clear that both the triangles ∆POT and QOT are right angled triangle.nd a common hypotenuse OT of these [ as shown in figure ]

Now, come to the concept , ∆POT and ∆QOT ∠OPT = OQT = 90° Common hypotenuse OT And OP = OQ [ OP and OQ are radii]So, R - H - S rule of similarity ∆POT ~ ∆QOT Hence, OP/OQ = PT/QT = OT/OT PT/QT = 1 PT = QT [ hence proved]

Theorem 9.2 : The lengths of tangents drawn froman external point to a circle are equalProof : We are given a circle with centre O, a pointlying outside the circle and two tangents PQ, PR on the circle from P (see Fig. 9.7). Were required to prove that PQ PR.Fig. 9.6

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