Mathematics - Circle Question with Solution | TestHub

$C_1\equiv x^2+y^2-2x-4y-4=0$..............(i)

And $C_2=x^2+y^2+2x+4y+4=0$ ...............(ii)

∴ Radical axis is $C_1-C_2=0$

$\Rightarrow -4x-8y-8=0$

Or $x+2y+2=0$ which is $L=0$

(a) option is correct

centre and radius of $C_1=0$ are $(1,2)$ and 3 .

∵ length of ⟂ from $(1,2)$ on $L=0$ is $\frac{|1+4+2|}{\sqrt{1+4}}=\frac{7}{\sqrt{5}}\ne$ radius

∴ (b) option is wrong $L$ is also the common chord of $C_1$ and $C_2$

∴ (c) option is correct. ∵ centres of $C_1=0$ and $C_2=0$ are ( 1,2 ) and ( $-1,-2$ )

∴ slope of line joining centres of circles $C_1=0\&C_2=0$ 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 $C_2$

∴ (d) option is correct