In fig 9.13 AB and CD are common tangents to two circles of unequal radii prove that AB = CD

Construct a line passing through AD

Now, AD and CD are tangents to the circle with centre O from the external point D. 
So, AD = CD    (Tangents drawn from an external point to a circle are equal.)         .....(1)

Also, AB and AD are the tangents to the circle with centre O' from the external point A.
So, AD = AB    (Tangents drawn from an external point to a circle are equal.)          .....(2)

From (1) and (2)

AB = CD

Hence Proved. 

As indicated by the inquiry,

Abdominal muscle = CD

Development: Produce AB and CD, to converge at P.

Confirmation:

Think about the circle with more noteworthy span.

Digressions drawn from an outside highlight a circle are equivalent

AP = CP … (1)

Too,

Think about the circle with more modest sweep.

Digressions drawn from an outer highlight a circle are equivalent

\[BP\text{ }=\text{ }BD\text{ }\ldots \text{ }\left( 2 \right)\]

Deduct Equation (2) from (1). We Get

\[AP\text{ }\text{ }BP\text{ }=\text{ }CP\text{ }\text{ }BD\]

Abdominal muscle = CD

Thus Proved.

Text Solution

Solution : Given AB and CD are common tangent to two circles of unequal radius <br> To prove AB=CD <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_MATH_X_C09_S01_025_S01.png" width="80%"> <br> Construction Produce AB and CD, to intersect at P. <br> Proof PA=PC <br> [the length of tangents drawn from an internal point to a circle are equal] <br> Also, PB=PD <br> [the lengths of tangents drawn from an internal point to a circle are equal] <br> `:.PA-PB=PC-PD` <br> AB=CD Hence proved.

>

Question 5In figure AB and CD are common tangents to two circles of unequal radii prove that AB = CD

Solution:

Given, AB and CD are the common tangents to two circles of unequal radii.

We have to prove that AB = CD.

Extend AB and CD such that it intersects at P.

We know that the tangents to a circle through an external point are equal.

Considering smaller circle,

The tangents are PB and PD

So, PB = PD ---------- (1)

Considering the larger circle,

The tangents are PA and PC

So, PA = PC ---------- (2)

Subtracting (1) and (2),

PA - PB = PC - PD

From the figure,

PA - PB = AB

PC - PD = CD

Therefore, AB = CD

✦ Try This: In the given figure, the length of tangents PA and PD are 8 cm and 3 cm respectively. Find the length of CD and AB.

Given, the length of tangent PA = 8 cm

The length of tangent PD = 3 cm

We have to find the length of CD and AB
We know that the tangents to a circle through an external point are equal.

So, PA = PC

Now, PC = 8 cm

Also, PD = PB

So, PB = 3 cm

From the figure,

AB = PA - PB

AB = 8 - 3 = 5 cm

CD = PC - PD

CD = 8 - 3 = 5 cm

Therefore, the length of AB and CD 5 cm and 5 cm.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10

NCERT Exemplar Class 10 Maths Exercise 9.3 Problem 5

Summary:

In Fig. 9.13, AB and CD are common tangents to two circles of unequal radii. It is proven that AB = CD

☛ Related Questions: