In fig. XP and XQ are tangents from X to the circle with center O. R is a point on the circle. Prove that, XA + AR = XB + BR. [2 MARKS]

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Solution

Concept: 1 Mark
Application: 1 Mark

Since lengths of tangents from an exterior point to a circle are equal.

$∴$ XP = XQ .... (i) [From X]

AP = AR .... (ii) [From A]

BQ = BR .... (iii) [From B]

Now, XP = XQ

$\Rightarrow X A + A P = X B + B Q$

Using equations (i) and (ii)

$\Rightarrow X A + A R = X B + B R$

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In fig. XP and XQ are tangents from X to the circle with center O. R is a point on the circle. Prove that, XA + AR = XB + BR. [2 MARKS]

Concept: 1 Mark Application: 1 Mark Since lengths of tangents from an exterior point to a circle are equal. ∴ XP = XQ .... (i) [From X] AP = AR .... (ii) [From A] BQ = BR .... (iii) [From B] Now, XP = XQ ⇒XA+AP=XB+BQ Using equations (i) and (ii) ⇒XA+AR=XB+BR