Question

Consider the circles
$C_1:x^2+y^2-2x-4y-4=0$ and
$C_2:x^2+y^2+2x+4y+4=0$
and the line $L:x+2y+2=0,$ then

• A. $L$ is the radical axis of $C_1$ and $C_2$
• B. $L$ is the common tangent of $C_1$ and $C_2$
• C. $L$ is the common chord of $C_1$ and $C_2$
• D. $L$ is the perpendicular to the line joining centres of $C_1$ and $C_2$

Answer 1

Answer: (A), (C), (D)

$L$ is the common chord of $C_1$ and $C_2$
$L$ is the perpendicular to the line joining centres of $C_1$ and $C_2$
$L$ is the radical axis of $C_1$ and $C_2$

Consider the circles
$C_1:x^2+y^2-2x-4y-4=0$ .........$\left(1\right)$
and $C_2:x^2+y^2+2x+4y+4=0$ ...........$\left(1\right)$
$\therefore$ Radical axis is $C_1-C_2=0$
$\Rightarrow -4x-8y-8=0$
or $x+2y+2=0$ which is $L=0$
$\left(a\right)$ option is correct.
Centre and radius of $C_1=0$ are $(1,2)$ and $3$
$\because$ Length of perpendicular from $(1,2)$ on $L=0$
is $\frac{|1+4+2|}{\sqrt{1+4}}=\frac{7}{\sqrt{5}}\ne$radius.
$\therefore$ option $\left(b\right)$ is wrong.
$L$ is also the common chord of $C_1$ and $C_2$
$\therefore$ option $\left(c\right)$ is correct.
$\because$ Centres of $C_1=0$ and $C_2=0$ are $(1,2)$ and $(-1,-2)$.
$\therefore$ Slope of line joining centres of circles $C_1=0$ and $C_2=0$ is $\frac{-2-2}{-1-1}=\frac{4}{2}=2=m_1$(say)
and Slope of $L=0$ is $-\frac{1}{2}=m_2$(say)
$\therefore m_1{m}_2=-1$
Hence $L$ is perpendicular to the line joining centres of $C_1$ and $C_2$
$\therefore \left(d\right)$ is correct.