♦ 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.

`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

smaller circle, is bisected at the point of contact.

Class 10 Chapter 10 Example 1

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

Prove that `∠ PTQ = 2 ∠ OPQ`.

Class 10 Chapter 10 Example 2

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

intersect at a point `T` (see Fig. 10.10). Find the length `TP`.

Class 10 Chapter 10 Example 3

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= 20/3`cm.

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`