First slide

Analysis of two circles in circles

Question

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

Difficult

A

$a^{2} = c^{2} (2 m + 1)$
B

$a^{2} = c^{2} (2 + m^{2})$
C

$c^{2} = a^{2} (2 + m^{2})$
D

$c^{2} = a^{2} (2 m + 1)$

Solution

Equation of family circles through (0, a) and $(0, - a)$ is
$[x^{2} + (y - a) (y + a)] + λ x = 0, λ \in R$
$\Rightarrow x^{2} + y^{2} + λ x - a^{2} = 0$
$\sqrt{{(\frac{λ}{2})}^{2} + a^{2}} = \frac{- \frac{m λ}{2} + c}{\sqrt{1 + m^{2}}}$
$\Rightarrow (1 + m^{2}) [\frac{λ^{2}}{4} + a^{2}] = {(\frac{m λ}{2} - c)}^{2}$
$\Rightarrow (1 + m^{2}) [\frac{λ^{2}}{4} + a^{2}] = \frac{m^{2} λ^{2}}{4} - m c λ + c^{2}$
$\Rightarrow λ^{2} + 4 m c λ + 4 a^{2} (1 + m^{2}) - 4 c^{2} = 0$
$∴ λ_{1} λ_{2} = 4 [a^{2} (1 + m^{2}) - c^{2}]$ since $λ$ is coefficient of x
$\Rightarrow g_{1} g_{2} = [a^{2} (1 + m^{2}) - c^{2}]$
since given the circles cut orthogonally
And $g_{1} g_{2} + f_{1} f_{2} = \frac{c_{1} + c_{2}}{2}$
$\Rightarrow a^{2} (1 + m^{2}) - c^{2} = - a^{2}$
Hence, $c^{2} = a^{2} (2 + m^{2})$

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