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(B) Let the equation of the circles be $x^2+y^2+2gx+2fy+d=0 \quad \ldots(1)$.Since these circles pass through $(0, a)$ and $(0, -a)$, we have $a^2+2fa

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$c^2=a^2(2+m^2)$

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(B) Let the equation of the circles be $x^2+y^2+2gx+2fy+d=0 \quad \ldots(1)$.Since these circles pass through $(0, a)$ and $(0, -a)$, we have $a^2+2fa+d=0 \quad \ldots(2)$ and $a^2-2fa+d=0 \quad \ldots(3)$.Solving $(2)$ and $(3)$, we get $f=0$ and $d=-a^2$.Substituting these values in $(1)$, we obtain $x^2+y^2+2gx-a^2=0 \quad \ldots(4)$.Since the line $y=mx+c$ touches this circle, the perpendicular distance from the centre $(-g, 0)$ to the line $mx-y+c=0$ is equal to the radius $\sqrt{g^2+a^2}$.Thus, $\frac{|-mg+c|}{\sqrt{1+m^2}} = \sqrt{g^2+a^2}$.Squaring both sides, we get $(c-mg)^2 = (1+m^2)(g^2+a^2)$.Expanding this, $c^2-2mcg+m^2g^2 = g^2+a^2+m^2g^2+m^2a^2$.Rearranging, $g^2 + 2mcg + a^2(1+m^2) - c^2 = 0$.Let $g_1$ and $g_2$ be the roots of this quadratic equation in $g$.Then the product of the roots is $g_1g_2 = a^2(1+m^2)-c^2 \quad \ldots(5)$.The two circles are $x^2+y^2+2g_1x-a^2=0$ and $x^2+y^2+2g_2x-a^2=0$.For these circles to cut orthogonally, $2g_1g_2 + 2f_1f_2 = d_1+d_2$.Here $f_1=f_2=0$ and $d_1=d_2=-a^2$, so $2g_1g_2 = -2a^2$, which implies $g_1g_2 = -a^2 \quad \ldots(6)$.Comparing $(5)$ and $(6)$, $-a^2 = a^2(1+m^2) - c^2$.Therefore, $c^2 = a^2(1+m^2) + a^2 = a^2(2+m^2)$.

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

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