The two circles which pass through (0, a) and | vedclass

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(C) The equation of any circle passing through $(0, a)$ and $(0, -a)$ is of the form $x^2 + y^2 + \lambda x - a^2 = 0$.This circle touches the line $m

Vedclass · Accepted Answer

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

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(C) The equation of any circle passing through $(0, a)$ and $(0, -a)$ is of the form $x^2 + y^2 + \lambda x - a^2 = 0$.This circle touches the line $mx - y + c = 0$. The perpendicular distance from the center $(-\lambda/2, 0)$ to the line is equal to the radius $r = \sqrt{(\lambda/2)^2 + a^2}$.Thus,$\frac{|m(-\lambda/2) - 0 + c|}{\sqrt{m^2 + 1}} = \sqrt{\frac{\lambda^2}{4} + a^2}$.Squaring both sides: $\frac{(c - m\lambda/2)^2}{m^2 + 1} = \frac{\lambda^2}{4} + a^2$.$(c - m\lambda/2)^2 = (m^2 + 1)(\frac{\lambda^2}{4} + a^2)$.$c^2 - mc\lambda + \frac{m^2\lambda^2}{4} = \frac{m^2\lambda^2}{4} + m^2a^2 + \frac{\lambda^2}{4} + a^2$.$\frac{\lambda^2}{4} + mc\lambda + (a^2(m^2 + 1) - c^2) = 0$.This is a quadratic in $\lambda$,giving two circles with parameters $\lambda_1$ and $\lambda_2$. For these circles to be orthogonal,$2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here,$g_1 = \lambda_1/2, g_2 = \lambda_2/2, f_1 = 0, f_2 = 0, c_1 = -a^2, c_2 = -a^2$.So,$2(\frac{\lambda_1}{2})(\frac{\lambda_2}{2}) = -a^2 - a^2$ $\Rightarrow \frac{\lambda_1\lambda_2}{2} = -2a^2$ $\Rightarrow \lambda_1\lambda_2 = -4a^2$.From the quadratic equation,the product of roots $\lambda_1\lambda_2 = 4(a^2(m^2 + 1) - c^2)$.Equating the two: $4(a^2(m^2 + 1) - c^2) = -4a^2$.$a^2(m^2 + 1) - c^2 = -a^2$.$c^2 = a^2(m^2 + 1 + 1) = a^2(m^2 + 2)$.

The two circles which pass through $(0, a)$ and $(0, -a)$ and touch the line $y = mx + c$ will intersect each other at a right angle,if

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