Mathematics

In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that :

(i) tangent at point P bisects AB.

(ii) angle APB = 90°.

In the figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that (i) tangent at point P bisects AB (ii) angle APB = 90°. Tangents and Intersecting Chords, Concise Mathematics Solutions ICSE Class 10.

Circles

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Answer

(i) We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

From figure,

In the figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that (i) tangent at point P bisects AB (ii) angle APB = 90°. Tangents and Intersecting Chords, Concise Mathematics Solutions ICSE Class 10.

TA and TP are the tangents to the circle with center O.

∴ TA = TP ………..(1)

TB and TP are the tangents to the circle with center O'.

∴ TB = TP ………..(2)

From (1) and (2) we get :

TA = TB.

Hence, proved that tangent at point P bisects AB.

(ii) In △ATP,

TA = TP [Proved above]

∴ ∠TAP = ∠TPA ………(1) [∵ angles opposite to equal sides are equal.]

In △BTP,

TB = TP [Proved above]

∴ ∠TBP = ∠TPB ……….(2) [∵ angles opposite to equal sides are equal.]

Adding (1) and (2), we get :

∠TAP + ∠TBP = ∠TPA + ∠TPB

∠TAP + ∠TBP = ∠APB ………..(3)

In △ABP,

⇒ ∠APB + ∠BAP + ∠ABP = 180° [Angle sum property of triangle]

⇒ ∠APB + ∠TAP + ∠TBP = 180° [From figure, ∠TAP = ∠BAP and ∠TBP = ∠ABP.]

⇒ ∠APB + ∠APB = 180°

⇒ 2∠APB = 180°

⇒ ∠APB = 90°.

Hence, proved that ∠APB = 90°.

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