Q92 / 160 · TS EAMCET 2019 (03 May Shift 1) · Mathematics

Each of the two orthogonal circles $C_1$ and $C_2$ passes through both the points $(2,0)$ and $(-2,0)$. If $y=m x+c$ is a common tangent to these circles, then

  • Correct answer
    A
    $c^2=4\left(1+2 m^2\right)$
  • B
    $c^2=2\left(1+2 m^2\right)$
  • C
    $c^2=1+m^2$
  • D
    $c^2 m^2=4\left(1+m^2\right)$

Solution

Let the equation of the circle be $x^2+y^2+2 g x+2 f y+c=0$ Since, this circle is passes through $(2,0)$ and $(-40)$ $$ \begin{aligned} \therefore \quad 4+4 g+c & =0, \\ 4-4 g+c & =0 \end{aligned} $$ On solving Eqs. (i) and (ii), we get $g=0, c=-4$ $\therefore$ Equation of circle $x^2+y^2+2 f y-4=0$ Now, $y=m x+c$ is tangent of circle $$ \begin{array}{ll} \therefore & \left|\frac{-f+c}{\sqrt{1+m^2}}\right|=\sqrt{f^2+4} \\ \Rightarrow & (c-f)^2=\left(1+m^2\right)\left(f^2+4\right) \\ \Rightarrow & m^2 f^2+2 c f+4\left(1+m^2\right)-c^2=0 \end{array} $$ Let $f_1 f_2$ are roots of equation $$ \therefore \quad f_1 f_2=\frac{4\left(1+m^2\right)-c^2}{m^2} $$ Now, equation of this circle are $$ x^2+y^2+2 f_1 y-4=0 \text { and } x^2+y^2+2 f_2 y-4=0 $$ Since, these are orthogonals. $$ \begin{aligned} & \therefore \quad 2 f_1 f_2=c_1+c_2 \\ & \Rightarrow \quad 2 f_1 f_2=-8 \\ & \Rightarrow \quad f_1 f_2=-4 \\ & \Rightarrow \frac{4\left(1+m^2\right)-c^2}{m^2}=-4 \\ & \Rightarrow \quad c^2=4\left(1+2 m^2\right) \\ & \end{aligned} $$

Hence, the correct answer is option: 'A'.

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