If ma and mb are two tangents drawn from a exterior point M to a circle with centre O

If PA and PB are two tangents drawn from a point P to a circle with center C touching it A and B, prove that CP is the perpendicular bisector of AB.

We shall prove that ∠ACP = ∠BCP = 90°and AC = BCNow, ∠APC = ∠BPCSince O lies on the bisector of ∠APB.

Δs ACP and BCP are congruent triangles by SAS congruence criterion,

Hence proved.

Concept: Tangent to a Circle

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From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the perpendicular bisector of AB.

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Question

From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the perpendicular bisector of AB.

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Correct option is A)

PA=PB

[length of tangents drawn from external point are equal ]

∠APC=∠BPC

[PA and PB are equally inclined to OP]

and PC=PC  [Common]

So, by SAS criteria of similarity

ΔPAC≅ΔPBC

⇒AC=BC and ∠ACP=∠BCP

But ∠ACP+∠BCP=180 ∘ ∴∠ACP+∠BCP=90 ∘ Hence, OP⊥AB

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If Pa and Pb Are Two Tangent Drawn from a Point P to a Circle with Centre C Touching It a and B, Prove that Cp is the Perpendicular Bisector of Ab.

If Pa and Pb Are Two Tangent Drawn from a Point P to a Circle with Centre C Touching It a and B, Prove that Cp is the Perpendicular Bisector of Ab.

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If PA and PB are two tangents drawn from a point P to a circle with center C touching it A and B, prove that CP is the perpendicular bisector of AB.

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SOLUTION

We shall prove that ∠ACP = ∠BCP = 90°

and AC = BC Now, ∠APC = ∠BPC

Since O lies on the bisector of ∠APB.

Δs ACP and BCP are congruent triangles by SAS congruence criterion,

∴ AC = BC and ∠ ACP = ∠ BCP

Since  ∠ ACP + ∠ BCP = 180°

2 ∠ ACP = 180° ∠ ACP = 90° ∠ACP = ∠BCP = 90° Hence proved.

Concept: Tangent to a Circle

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Chapter 15: Circles - Exercise

Q 25 Q 24 Q 26

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ICSE Class 10 Mathematics

Chapter 15 Circles Exercise | Q 25 VIEW ALL [4] view

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If PA and PB are two tangents drawn from a point P to a circle with centre O touching it at A and B prove that OP is perpendicular bisector of AB

May 02,2022 - If PA and PB are two tangents drawn from a point P to a circle with centre O touching it at A and B prove that OP is perpendicular bisector of AB | EduRev Class 10 Question is disucussed on EduRev Study Group by 1643 Class 10 Students.

Class 10 Question

If PA and PB are two tangents drawn from a point P to a circle with centre O touching it at A and B prove that OP is perpendicular bisector of AB

Answers

Manish Singh Apr 06, 2021

Upvote | 7 Reply Nirmal Kumar Dec 19, 2019

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Prashant Dwivedi Oct 22, 2017

Join OBWe know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.angle OAP = angle OBP = 90�Now,angle OAP + angle APB + angle OBP + angle AOB = 360� [Angle sum property of quadrilaterals]implies 90�+ angle APB + 90� + angle AOB = 360�implies angle AOB = 360� - 180� - angle APB = 180� - angle APB ....(1)Now, in triangle OAB, OA is equal to OB as both are radii.implies angle OAB = angle OBA [In a triangle, angles opposite to equal sides are equal]Now, on applying angle sum property of triangles in Triangle AOB, we obtainAngle OAB + angle OBA + angle AOB = 180�implies 2angle OAB + angle AOB = 180�implies 2 angle OAB + (180� - angle APB) = 180� [Using (1)]implies 2 angle OAB = angle APBThus, the given result is proved

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