The two circles which passes through (0,a) an | vedclass

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Question

(c) Equation of circles

$[{x^2} + (y - a)(y + a)] + \lambda x = 0 \Rightarrow {x^2} + {y^2} + \lambda x - {a^2} = 0$

and $\sqrt {{{\left( {\frac

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) Equation of circles

$[{x^2} + (y - a)(y + a)] + \lambda x = 0 \Rightarrow {x^2} + {y^2} + \lambda x - {a^2} = 0$

and $\sqrt {{{\left( {\frac{\lambda }{2}} ight)}^2} + {a^2}} = \frac{{\frac{{ - m\lambda }}{2} + c}}{{\sqrt {1 + {m^2}} }}$

$ \Rightarrow (1 + {m^2}){m{ }}\left[ {\frac{{{\lambda ^2}}}{4} + {a^2}} ight] $

$= {\left( {\frac{{m\lambda }}{2} - c} ight)^2}$

$ \Rightarrow (1 + {m^2}){m{ }}\left[ {\frac{{{\lambda ^2}}}{4} + {a^2}} ight] $

$= \frac{{{m^2}{\lambda ^2}}}{4} - mc\lambda + {c^2}$

$ \Rightarrow {\lambda ^2} + 4mc\lambda + 4{a^2}(1 + {m^2}) - 4{c^2} = 0$

$	herefore $${\lambda _1}{\lambda _2} = 4[{a^2}(1 + {m^2}) - {c^2}]$

$\Rightarrow {g_1}{g_2} = [{a^2}(1 + {m^2}) - {c^2}]$

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 ${c^2} = {a^2}(2 + {m^2})$.

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

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