Question

$C_1=x^2+y^2-4x-8y-5=0$ and $C_2=x^2+y^2-12x-2y+12=0$ are the equations of two given circles. The length of the common chord of $C_1$ and $C_2$ is:

• A. $2\sqrt{3}$
• B. $\frac{5\sqrt{3}}{2}$
• C. $5\sqrt{3}$
• D. $4\sqrt{3}$

Answer 1

Answer: (C)

$5\sqrt{3}$

Let $A$ and $C$ be centres of the given circles $C_1$ and $C_2$
]

$C_1:x^2+y^2-4x-8y-5=0$

$r_1=\sqrt{2^2+4^2+5}=5$

$C_2:x^2+y^2-12x-2y+12=0$
$r_2=\sqrt{6^2+1^2-12}=5$

$AC=\sqrt{(6-2{)}^2+(1-4{)}^2}=5$

Hence, $A$ lies on $C_2$ and $C$ lies on $C_1$

Hence, $\Delta ABC$ is an equilateral triangle.

The length of the common chord $BD$ will be twice the length of the altitude of $\Delta ABC$.
Hence,

$BD=2\times 5sin\frac{\pi }{3}=2\times 5\frac{\sqrt{3}}{2}=5\sqrt{3}$
Hence, option C is correct.