Question

The circles $x^2+y^2+2g_{1}x-a^2=0$ and $x^2+y^2+2g_{2}x-a^2=0$ cut each other orthogonally. If $p_1,p_2$ are perpendicular from $(0,a)$ and $(0,-a)$ on a common tangent of these circles the $p_1,p_2=$

• A. $3a^2$
• B. $a^2$
• C. $2a^2$
• D. $a^2+2$

Answer 1

Answer: (B)

$a^2$

Since the given circles cut each other orthogonally $g_1{g}_2+a^2=0$ ...(1)

If $lx+my=1$ is common tangent of these circles, then

$\frac{-lg-1}{\sqrt{l^2+m^2}}=\pm \sqrt{{g_1}^2+a^2}\Rightarrow {(l{g}_1+1)}^2=(l^2+m^2)({g_1}^2+a^2)\Rightarrow m^2{g_1}^2-2l{g}_1+a^2(l^2+m^2)-1=0$

Similarly $m^2{g_2}^2-2l{g}_2+a^2(l^2+m^2)-1=0$

Thus $g_1,g_2$ are the roots of the equation

$m^2{g}^2-2lg+a^2(l^2+m^2)-1=0$

$\Rightarrow g_1{g}_2=\frac{a^2(l^2+m^2)-1}{m^2}=-a^2$ (from (1))

$\Rightarrow a^2(l^2+m^2)=1-a^2{m}^2$ ...(2)

Now $p_1{p}_2=\frac{|ma-1|}{\sqrt{l^2+m^2}}.\frac{|-ma-1|}{\sqrt{l^2+m^2}}=\frac{|1-m^2{a}^2|}{l^2+m^2}=a^2$ (by (2)