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STD 11 - 10.1 circle and system of circle — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 10.1 circle and system of circle4 Marks
MCQ
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
  • A
    ${a^2} = {c^2}(2m + 1)$
  • B
    ${a^2} = {c^2}(2 + {m^2})$
  • ✓
    ${c^2} = {a^2}(2 + {m^2})$
  • D
    ${c^2} = {a^2}(2m + 1)$
✓

Answer

Correct option: C.
${c^2} = {a^2}(2 + {m^2})$
c
(c) Equation of circles

$[{x^2} + (y - a)(y + a)] + \lambda x = 0 \Rightarrow {x^2} + {y^2} + \lambda x - {a^2} = 0$

and $\sqrt {{{\left( {\frac{\lambda }{2}} \right)}^2} + {a^2}} = \frac{{\frac{{ - m\lambda }}{2} + c}}{{\sqrt {1 + {m^2}} }}$

$ \Rightarrow (1 + {m^2}){\rm{ }}\left[ {\frac{{{\lambda ^2}}}{4} + {a^2}} \right] $

$= {\left( {\frac{{m\lambda }}{2} - c} \right)^2}$

$ \Rightarrow (1 + {m^2}){\rm{ }}\left[ {\frac{{{\lambda ^2}}}{4} + {a^2}} \right] $

$= \frac{{{m^2}{\lambda ^2}}}{4} - mc\lambda + {c^2}$

$ \Rightarrow {\lambda ^2} + 4mc\lambda + 4{a^2}(1 + {m^2}) - 4{c^2} = 0$

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

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