Q49 / 160 · AP EAMCET 2014 · Mathematics
The length of the common chord of the two circles $x^2+y^2-4 y=0$ and $x^2+y^2-8 x$ $-4 y+11=0$, is
- A$\frac{\sqrt{145}}{4} \mathrm{~cm}$
- B$\frac{\sqrt{11}}{2} \mathrm{~cm}$
- C$\sqrt{135} \mathrm{~cm}$
- Correct answerD$\frac{\sqrt{135}}{4} \mathrm{~cm}$
Solution
Given equation of circles are
and
$$
\begin{gathered}
x^2+y^2-4 y=0 \\
x^2+y^2-8 x-4 y+11=0
\end{gathered}
$$
$\therefore$ Equation of chord
$$
\begin{aligned}
& \quad x^2+y^2-4 y-\left(x^2+y^2-8 x-4 y+11\right)=0 \\
& \Rightarrow \quad 8 x-11=0
\end{aligned}
$$
Centre and radius of first circle are $O(0,2)$
and
$$
O P=r=2 .
$$
Now, perpendicular distance from $O(0,2)$ to the line $8 x-11$ is
$$
d=O M=\frac{|8 \times 0-11|}{\sqrt{8^2}}=\frac{11}{8}
$$
In $\triangle O M P$,
$$
\begin{aligned}
P M & =\sqrt{O P^2-O M^2} \\
& =\sqrt{2^2-\left(\frac{11}{8}\right)^2} \\
& =\sqrt{4-\frac{121}{64}}=\sqrt{\frac{256-121}{64}} \\
& =\frac{\sqrt{135}}{8}
\end{aligned}
$$
$\therefore$ Length of chord $P Q=2 P M=2 \times \frac{\sqrt{135}}{8}$
$$
=\frac{\sqrt{135}}{4} \mathrm{~cm}
$$
Hence, the correct answer is option: 'D'.