Question

The length of the common chord of the two circles $x^2+y^2-4y=0$ and $x^2+y^2-8x-4y+11=0$, 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
$x^2+y^2-4y=0$
and $x^2+y^2-8x-4y+11=0$
$\therefore$ Equation of chord
$x^2+y^2-4y-\left(x^2+y^2-8x-4y+11\right)=0$
$\Rightarrow 8x-11=0$
Centre and radius of first circle are $O(0,2)$ and $OP=r=2$.
Now, perpendicular distance from $O(0,2)$ to the line $8x-11$ is
$d=OM=\frac{|8\times 0-11|}{\sqrt{8^2}}=\frac{11}{8}$
In $△OMP$,
$PM=\sqrt{O{P}^2-O{M}^2}$
$=\sqrt{2^2-{\left(\frac{11}{8}\right)}^2}$
$=\sqrt{4-\frac{121}{64}}=\sqrt{\frac{256-121}{64}}$
$=\frac{\sqrt{135}}{8}$
$\therefore$ Length of chord $PQ=2PM=2\times \frac{\sqrt{135}}{8}$
$=\frac{\sqrt{135}}{4}cm$