Question

Two circles of radii $3,4$ intersect orthogonally. Then the length of the common chord is

• A. $\frac{12}{5}$
• B. $\frac{24}{25}$
• C. $\frac{24}{5}$
• D. $\frac{25}{24}$

Answer 1

Answer: (C)

$\frac{24}{5}$

L$=$length of common chord
Given : Two circles of radii $3,4$ intersect orthogonally
So, $\angle Q={90}^0$
$\therefore (C_1{C}_2{)}^2=3^2+4^2$
$\therefore \left(C_1{C}_2\right)=5$
$\therefore A\left(\Delta A{C}_1{C}_2\right)=\frac{1}{2}\times 4\times 3=6....\left(1\right)$
And also, $AB$ is perpendicular to $C_1{C}_2$
Area of $\Delta A{C}_1{C}_2=\frac{1}{2}\times C_1{C}_2\times \frac{L}{2}$
$=\frac{1}{4}\times 5\times L.....\left(2\right)$
From (1) & (2),
$\Rightarrow \frac{5L}{4}=6$
$L=\frac{24}{5}$