The length of the common chord of the circles of radii 15 and 20, whose centres are 25 unit of distance…
- 12
- 16
- 24
- 25
Solution
Given, $r_1=15$ unit $r_2=20 \text { unit }$ $C_1 C_2=25$ unit Let $\angle A C_2 D=\theta$ Then, in right angled triangle $A D C_2$, $A D=r_2 \sin \theta$ $\Rightarrow \quad \frac{A D}{r_2}=\sin \theta$ ...(i) Now, in right angled triangle $A D C_1$ $A D=r_1 \sin (90-\theta)$ $\Rightarrow \quad \frac{A D}{r_1}=\cos \theta$ ...(ii) On squaring and adding Eqs. (i) and (ii), we get $A D^2\left[\frac{1}{r_1^2}+\frac{1}{r_2^2}\right]=1$ $\Rightarrow \quad A D^2\left(\frac{r_1^2+r_2^2}{r_1^2 r_2^2}\right)=1$ $\Rightarrow \quad A D^2=\frac{r_1^2 r_2^2}{r_1^2+r_2^2}$ $\Rightarrow \quad A D^2=\frac{225 \times 400}{225+400}$ $\Rightarrow \quad A D=\frac{15 \times 20}{25}=12$ Thus, length of common chord $=2 A D$ $=2 \times 12$ $=24$ unit
Asked in: AP EAMCET 2010