Question

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. $\frac{\sqrt{145}}{4} \,cm$
• B. $\frac{\sqrt{11}}{2} \,cm$
• C. $\sqrt{135}\, cm$
• D. $\frac{\sqrt{135}}{4}\, cm$

Answer 1

Answer: (D)

Given equation of circles are
$x^{2}+y^{2}-4 y=0$
and $x^{2}+y^{2}-8 x-4 y+11=0$
$\therefore $ Equation of chord
$x^{2}+y^{2}-4 y-\left(x^{2}+y^{2}-8 x-4 y+11\right)=0$
$\Rightarrow 8 x-11=0$
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 $\triangle OMP$,
$ 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} $
$\therefore $ Length of chord $P Q=2 P M=2 \times \frac{\sqrt{135}}{8}$
$=\frac{\sqrt{135}}{4} \,cm$