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Tangents to a Circle

Summary

LearnNext Lesson Video

HD 7:21 Animated video Lecture for Tangents to a Circle

A straight line intersects a circle at one or two point. The tangent to a circle is a line that touches the circle at one point. Secant intersects the circle at two points. The point at which the straight line touches the circle is called the point of contact or point of tangency.
Some properties of tangents to a circle are Infinite number of tangents can be drawn to a circle but only one tangent can be drawn at any given point on a circle.  From an external point we can draw two tangents of equal length. The radius of the circle is perpendicular to the tangent at its point of contact and the tangents drawn at the extremities of the diameter of a circle are parallel.

 

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The tangent at any point on a circle is perpendicular to the radius drawn to the point of contact.

Given: A tangent AB with point of contact P.

To prove: OP ⊥ AB

Proof:

Consider point C on AB other than P.

C must lie outside the circle. (∵ A tangent can have only one point of contact with the circle)

OC > OP (∵ C lies outside the circle)

This is true for all positions of C on AB.

Thus, OP is the shortest distance between point P and line segment AB.

Hence, OP ⊥ AB.

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From an external point we can draw only two tangents to a circle.


Theorem: Tangents drawn to a circle from an external point are equal in length.

Given: Two tangents AB and AC from an external point A to points B and C on a circle.

To prove: AB = AC

Construction: Join OA, OB and OC.

Proof:

In triangles OAB and OAC,

∠OBA = 90o (Radius OB ⊥ Tangent AB at B)

∠OCA = 90o (Radius OC ⊥ Tangent AC at C)

In triangles OBA and OCA,

∠OBA = ∠OCA = 90o

OB = OC (Radii of the same circle)

OA = OA (Common side)

Thus, ΔOBA ΔOCA (RHS congruence rule)

Hence, AB = AC (Corresponding sides of congruent triangles)

The tangents drawn at the extremities of the diameter of a circle are parallel.

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Questions & Answers

1 . Prove that any four vertices of a regular pentagon are concyclic

Given ABCDE is a regular pentagon
That is AB = BC = CD = DE = AE
Recall that the sum of angles in a regular pentagon...
2 . Perimeter of a right triangle is equal to the sum of the diameter

ABC is a right angled triangle right angled at B.
Hence AC is diameter of the circumcircle as angle in a semi circle is a rig...
3 . ABC is right angle with angel B=90".

Since BC is diameter of the circle, ∠BDC = 90° [Since angle in a semi circle is a right angle]
Consider Δ’s ABC and ADB...
4 . form an external point P tangent PA and PB are drawn to a circle .If CD is the tangent to the circle at a point E and PA=15cm,find the perimeter of the triangle PCD.
Sol :

Perimeter of ΔPCD = 2 x AP.
                            =2 x 15 = 30 cm.
5 . Prove that ΔPRS ~ΔPTQ. If PQ = 4cm ,PT=3cm , ST= 5cm.
Sol:
PQ = 4 cm, PT = 3 cm , ST = 5 cm
(i)
PQ x PR = PT x PS
4 x PR = 3 x 8
PR = 6 cm
QR = PR...