Class 10 Number of Tangents from a Point on a Circle

### Topic Covered

♦ Number of Tangents from a Point on a Circle

### Number of Tangents from a Point on a Circle

To get an idea of the number of tangents from a point on a circle, let us perform the following activity:

text ( Activity 3 :) Draw a circle on a paper. Take a point P inside it. Here, it is not possible to draw any tangent to a circle through a point inside it [see Fig. 10.6 (i)].

Next take a point P on the circle and draw tangents through this point. You have already observed that there is only one tangent to the circle at such a point [see Fig. 10.6 (ii)].

Finally, take a point P outside the circle and try to draw tangents to the circle from this point.

Now, You will find that you can draw exactly two tangents to the circle through this point [see Fig. 10.6 (iii)].

We can summarise these facts as follows:

text ( Case 1 : ) There is no tangent to a circle passing through a point lying inside the circle.

text (Case 2 :) There is one and only one tangent to a circle passing through a point lying on the circle.

text ( Case 3 : ) There are exactly two tangents to a circle through a point lying outside the circle.

In Fig. 10.6 (iii), T_1 and T_2 are the points of contact of the tangents PT_1 and PT_2 respectively.

The length of the segment of the tangent from the external point P and the point of contact with the circle is called the length of the tangent from the point P to the circle.

Note that in Fig. 10.6 (iii), PT_1 and PT_2 are the lengths of the tangents from P to the circle. The lengths PT_1 and PT_2 have a common property. Can you find this? Measure PT_1 and PT_2.

In fact, this is always so. Let us give a proof of this fact in the following theorem.

text (Theorem 10.2 : ) The lengths of tangents drawn from an external point to a circle are equal.

text ( Proof :)

We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P (see Fig. 10.7). We are required to prove that PQ = PR.

For this, we join OP, OQ and OR. Then ∠ OQP and ∠ ORP are right angles, because these are angles between the radii and tangents, and according to Theorem 10.1 they are right angles. Now in right triangles OQP and ORP,

OQ = OR (Radii of the same circle)

OP = OP (Common)

Therefore, Delta OQP ≅ Delta ORP (RHS)

This gives PQ = PR (CPCT)

text (Remarks : )

1. The theorem can also be proved by using the Pythagoras Theorem as follows:

PQ^2 = OP^2 – OQ^2 = OP^2 – OR^2 = PR^2 (As OQ = OR)

which gives PQ = PR.

2. Note also that ∠ OPQ = ∠ OPR. Therefore, OP is the angle bisector of ∠ QPR,
i.e., the centre lies on the bisector of the angle between the two tangents.
Q 3129491311

Prove that in two concentric circles, the chord of the larger circle, which touches the
smaller circle, is bisected at the point of contact.
Class 10 Chapter 10 Example 1
Solution:

We are given two concentric circles

C_1 and C_2 with centre O and a chord AB of the
larger circle C_1 which touches the smaller circle
C_2 at the point P (see Fig. 10.8). We need to prove
that AP = BP.

Let us join OP. Then, AB is a tangent to C_2 at P
and OP is its radius. Therefore, by Theorem 10.1,

OP ⊥ AB

Now AB is a chord of the circle C_1 and OP ⊥ AB. Therefore, OP is the bisector of the
chord AB, as the perpendicular from the centre bisects the chord,

i.e., AP = BP
Q 3159491314

Two tangents TP and TQ are drawn to a circle with centre O from an external point T.
Prove that ∠ PTQ = 2 ∠ OPQ.
Class 10 Chapter 10 Example 2
Solution:

We are given a circle with centre O, an external point T and two tangents TP and TQ
to the circle, where P, Q are the points of contact (see Fig. 10.9). We need to prove that

∠ PTQ = 2 ∠ OPQ

Let ∠ PTQ = θ

Now, by Theorem 10.2, TP = TQ. So, TPQ is an isosceles triangle.

Therefore,  ∠ TPQ = ∠ TQP = 1/2 (180^o -θ ) = 90^o -1/2 θ

Also, by Theorem 10.1, ∠ OPT = 90°

So, ∠ OPQ = ∠ OPT – ∠ TPQ = 90^o - (90^o -1/2 θ )

=1/2 θ = 1/2 ∠ PTQ

This gives ∠ PTQ = 2 ∠ OPQ
Q 3109491318

PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q
intersect at a point T (see Fig. 10.10). Find the length TP.
Class 10 Chapter 10 Example 3
Solution:

Join OT. Let it intersect PQ at the point R. Then Delta TPQ is isosceles and TO is the
angle bisector of ∠ PTQ. So, OT ⊥ PQ and therefore, OT bisects PQ which gives

PR = RQ = 4 cm.

Also, OR = sqrt (OP^2 -PR^2 ) = sqrt (5^2 - 4^2) cm = 3 cm .

Now, ∠ TPR + ∠ RPO = 90° = ∠ TPR + ∠ PTR (Why?)

So, ∠ RPO = ∠ PTR

Therefore, right triangle TRP is similar to the right triangle PRO by AA similarity.

This gives  (TP)/(PO) = (RP)/(RO), i.e.,  (TP)/5 = 4/3 or TP= 20/3cm.

Note : TP can also be found by using the Pythagoras Theorem, as follows:

Let TP = x and TR = y. Then

x^2 = y^2 + 16 (Taking right Delta PRT) .........(1)

x^2 + 5^2 = (y + 3)^2 (Taking right Delta OPT).......... (2)

Subtracting (1) from (2), we get

25 = 6y – 7 or y = (32)/6 =16/3

Therefore, x^2 = (16/3)^2 +16 =16/9 (16 +9 ) = (16 xx 25)/9 [From (1)]

or x = 20/3