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

Answer 1

The correct option is D $\frac{\sqrt{135}}{4}$

$s_1=x^2+y^2-4y=0$
$s_2=x^2+y^2-8x-4y+11=0$
Equation common chord $=s_1-s_2=8x-11$
$OC=\left|\frac{-11}{\sqrt{8^2}}\right|=\frac{11}{8}$
$CD=\left|\frac{32-11}{\sqrt{8^2}}\right|=\left|\frac{21}{8}\right|$
$A{C}^2=O{A}^2-O{C}^2$
$=2^2-{\left(\frac{11}{8}\right)}^2$
$=\frac{256-121}{64}$
$=\frac{135}{64}$
$AC=\sqrt{\frac{135}{64}}=\sqrt{\frac{135}{8}}$
$2AC=\frac{\sqrt{135}}{4}$ is the length of common chord.