Question

Two circles $S_1$ and $S_2$ pass through the points $(0,a)$ and $(0,-a)$. The line $y=mx+c$ is a tangent to the two circles. If $S_1$ and $S_2$ are orthogonal, then

• A. $a^2(2m^2+1)=c^2$
• B. $a^2(m^2+2)=c^2$
• C. $c^2(m^2+1)=a^2$
• D. $c^2(m^2+2)=a^2$

Answer 1

The correct option is A $a^2(2m^2+1)=c^2$

Consider (0,a) and (0,-a) the centres of circle $S_1$ & $S_2$ respectively otherwise the situation will be complex & $S_1$ & $S_2$ are orthogonal then $2g{g}_1+2f{f}_1=d+d_1$, condition for orthogonality

${(x-0)}^2+{(y-a)}^2={r_1}^2\Rightarrow$radius of $S_1$

${(x-0)}^2+{(y+a)}^2={r_2}^2\Rightarrow$radius of $S_2$

General formula of circle

$x^2+y^2+2gx+2fy+d=0$

$x^2+y^2-2ay+a^2-{r_1}^2=0$

$x^2+y^2+2ay+a^2-{r_2}^2=0$

$d=a^2-{r_1}^2,2g{g}_1+2f{f}_1=d+d_1$

$d_1=a^2-{r_2}^2,g=0=g_1,f=a,f_1=-a$

So, $-2a^2=a^2-{r_1}^2+a^2-{r_2}^2$

$4a^2={r_1}^2+{r_2}^2⟶1$

$mx+c-y=0$ tangent's equation

Now on circle $S_1(0,a)$

$\frac{m\times 0+c-a}{\sqrt{m^2+1}}=r_1$

On circle $S_2$

$\frac{m\times 0+c+a}{\sqrt{m^2+1}}=r_2$

Putting $r_1\&r_2$'s value in equation 1

$4a^2=\frac{{(c-a)}^2+{(c+a)}^2}{m^2+1}$

$4a^2=\frac{c^2+a^2-2ac+c^2+a^2+2ac}{(m^2+1)}$

$\Rightarrow 2a^2(m^2+1)=c^2+a^2$

$\Rightarrow a^2(2m^2+1)=c^2$