Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm.
Question.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Solution:
Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively.
$O A=O B=5 \mathrm{~cm}$
$O^{\prime} A=O^{\prime} B=3 \mathrm{~cm}$
OO' will be the perpendicular bisector of chord $A B$.
$\therefore A C=C B$
It is given that, $O O^{\prime}=4 \mathrm{~cm}$
Let $O C$ be $x$. Therefore, $O^{\prime} C$ will be $x-4$
In $\triangle O A C$
$O A^{2}=A C^{2}+O C^{2}$
$\Rightarrow 5^{2}=A C^{2}+x^{2}$
$\Rightarrow 25-x^{2}=A C^{2} \ldots(1)$
In $\Delta \mathrm{O}^{\prime} \mathrm{AC}$,
$O^{\prime} A^{2}=A C^{2}+O^{\prime} C^{2}$
$\Rightarrow 3^{2}=\mathrm{AC}^{2}+(x-4)^{2}$
$\Rightarrow 9=A C^{2}+x^{2}+16-8 x$
$\Rightarrow \mathrm{AC}^{2}=-x^{2}-7+8 x \ldots(2)$
From equations (1) and (2), we obtain
$25-x^{2}=-x^{2}-7+8 x$
$8 x=32$
$x=4$
Therefore, the common chord will pass through the centre of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle.
Length of the common chord $A B=2 O^{\prime} A=(2 \times 3) \mathrm{cm}=6 \mathrm{~cm}$
Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively.
$O A=O B=5 \mathrm{~cm}$
$O^{\prime} A=O^{\prime} B=3 \mathrm{~cm}$
OO' will be the perpendicular bisector of chord $A B$.
$\therefore A C=C B$
It is given that, $O O^{\prime}=4 \mathrm{~cm}$
Let $O C$ be $x$. Therefore, $O^{\prime} C$ will be $x-4$
In $\triangle O A C$
$O A^{2}=A C^{2}+O C^{2}$
$\Rightarrow 5^{2}=A C^{2}+x^{2}$
$\Rightarrow 25-x^{2}=A C^{2} \ldots(1)$
In $\Delta \mathrm{O}^{\prime} \mathrm{AC}$,
$O^{\prime} A^{2}=A C^{2}+O^{\prime} C^{2}$
$\Rightarrow 3^{2}=\mathrm{AC}^{2}+(x-4)^{2}$
$\Rightarrow 9=A C^{2}+x^{2}+16-8 x$
$\Rightarrow \mathrm{AC}^{2}=-x^{2}-7+8 x \ldots(2)$
From equations (1) and (2), we obtain
$25-x^{2}=-x^{2}-7+8 x$
$8 x=32$
$x=4$
Therefore, the common chord will pass through the centre of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle.
Length of the common chord $A B=2 O^{\prime} A=(2 \times 3) \mathrm{cm}=6 \mathrm{~cm}$
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