Question

The external centre of similitude of the circles $x^2+y^2-12x+8y+48=0$ and $x^2+y^2-4x+2y-4=0$ is

• A. $(14,10)$
• B. $(-14,-10)$
• C. $(-14,10)$
• D. $(14,-10)$

Answer 1

The correct option is D $(14,-10)$

Given circles are
$x^2+y^2-12x+8y+48=0$ and $x^2+y^2-4x+2y-4=0$
The centre and radius are
$C_1=(6,-4),r_1=2$
$C_2=(2,-1),r_2=3$


Here, $P$ is the external centre of simlitude divides $C_1$ and $C_2$ in ratio of $r_1:r_2$ externally
$\Rightarrow \frac{P{C}_1}{P{C}_2}=2:3$ externally
So,
$P=(\frac{2\left(2\right)-3\left(6\right)}{2-3},\frac{2(-1)-3(-4)}{2-3})\therefore P=(14,-10)$