Name: Proof of Angle in the Same Segment of a Circle
Uploaded: 2019-07-19
Duration: 1 min 26 s
Description: The most important and simple theorem from the chapter circle that angle in the same segment of the circle are equal. Let's derive this theorem.

Question

Two circles of radii

5 c m

and

3 c m

intersect at two points and the distance between their Centres is

4 c m .

Find the length of the common chord.

Accepted Answer

Radius of circles centred at $O$ and $3 c m$ be $5 c m$ and $3 c m$ respectively.

$O A = O B = 5 c m$

$O^{'} A = O^{'} B = 3 c m$

$O O'$ will be the perpendicular bisector of chord $A B .$

$∴ A C = C B$

Given that $O O^{'} = 4 c m$

Let $O C = x .$

So, $O^{'} C = (4 - x) cm$

In $△ O A C$ ,

${O A}^{2} = {A C}^{2} + {O C}^{2}$ (using Pythagoras theorem)

$\Rightarrow 5^{2} = {A C}^{2} + x^{2}$

$\Rightarrow 25 - x^{2} = {A C}^{2}$ $. . . (1)$

In $△ O' A C$ ,

$O^{'} A^{2} = {A C}^{2} + O^{'} C^{2}$ (using Pythagoras theorem)

$\Rightarrow 3^{2} = {A C}^{2} + {(4 - x)}^{2}$

$\Rightarrow 9 = {A C}^{2} + 16 + x^{2} - 8 x$

$\Rightarrow {A C}^{2} = - x^{2} - 7 + 8 x$ $. . . (2)$

From equations $(1)$ and $(2)$ ,

$25 - x^{2} = - x^{2} - 7 + 8 x$

$\Rightarrow 8 x = 32$

$\Rightarrow x = 4 cm$ .

$\Rightarrow O' C = 4 - 4 = 0$

So, the common chord will pass through the centre of smaller circle $O'$ and hence it will be diameter of smaller circle.

Now, ${A C}^{2} = 25 - x^{2}$ [from equation $(1)$ ]

$= 25 - 4^{2}$

$= 25 - 16$

$= 9$

$∴ A C = 3 c m$

The length of the common chord $A B = 2 A C = (2 \times 3) c m = 6 c m .$

EXERCISE

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