Question

If the two circles, which pass through $(0,a)$ and $(0,-a)$ and touch the line $y=mx+c$, will cut orthogonally, if

• A. $c^2=a^2(2-m^2)$
• B. $\frac{a^2}{c^2}=\frac{1}{2+m^2}$
• C. $c^2=a^2(2+m^2)$.
• D. $c^2=a^2(1-m^2)$.

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{a^2}{c^2}=\frac{1}{2+m^2}$
C $c^2=a^2(2+m^2)$.

Let the equation of the circles be
$x^2+y^2+2gx+2fy+d=0\dots \left(1\right)$

Since, these circles pass through $(0,a)$ and $(0,-a)$ then
$a^2+2fa+d=0\dots \left(2\right)$
and $a^2-2fa+d=0\dots \dots \left(3\right)$
Solving $\left(2\right)$ and $\left(3\right),$ we get
$f=0$ and $d=-a^2$.

Substituting these values of $f$ and $d$ in $\left(1\right)$, we obtain

$x^2+y^2+2gx-a^2=0\dots \dots \left(4\right)$

Now, $y=mx+c$ touch this circle.
Therefore, perpendicular distance from the centre = radius
$\Rightarrow \frac{|-mg-0+c|}{\sqrt{1+m^2}}=\sqrt{(g^2+a^2)}$
Squaring both the sides
$\Rightarrow (c-mg{)}^2=(1+m^2\left)\right(g^2+a^2)$
$\Rightarrow g^2+2mcg+a^2(1+m^2)-c^2=0$

Let $g_1,g_2$ are the roots of this equation

$g_1{g}_2=a^2(1+m^2)-c^2\dots \left(5\right)$

Now, the equation of the two circles represented by $\left(4\right)$ are

$x^2+y^2+2g_{1}x-a^2=0$ and $x^2+y^2+2g_{2}x-a^2=0$

These two circles will cut orthogonal, if

$2g_1{g}_2+0=-a^2-a^2$
$\Rightarrow g_1{g}_2=-a^2\dots \left(6\right)$

From $\left(5\right)$ and $\left(6\right),$
$-a^2=a^2(1+m^2)-c^2$
$\Rightarrow c^2=a^2(2+m^2)$