Q94 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics

If $(h, k)$ is the centre of the circle which passes through the origin and cuts the circles $x^2+y^2+4 x+6 y+12=0$ and $x^2+y^2+4 x-6 y+9=0$ orthogonally, then $k-2 h=$

  • A
    0
  • B
    1
  • C
    -1
  • Correct answer
    D
    5

Solution

We have the equation of circle passes through the origin and cutting the circles $x^2+y^2-2 g_1 x-2 f_1 y+c_1=0$ and $x^2+y^2-2 g_2 x-2 f_2 y+c_2=0$ orthogonally is $$ \left|\begin{array}{ccc} \mathrm{x}^2+\mathrm{y}^2 & \mathrm{x} & \mathrm{y} \\ \mathrm{c}_1 & \mathrm{~g}_1 & \mathrm{f}_1 \\ \mathrm{c}_2 & \mathrm{~g}_2 & \mathrm{f}_2 \end{array}\right|=0 $$ $\therefore$ Equation of required circle is $$ \left|\begin{array}{ccc} x^2+y^2 & x & y \\ 12 & -2 & -3 \\ 9 & -2 & 3 \end{array}\right|=0 $$ $$ \begin{aligned} & \Rightarrow\left(\mathrm{x}^2+\mathrm{y}^2\right)(-6-6)-\mathrm{x}(36+27)+\mathrm{y}(-24+18)=0 \\ & \Rightarrow-12\left(\mathrm{x}^2+\mathrm{y}^2\right)-63 \mathrm{x}-6 \mathrm{y}=0 \\ & \Rightarrow \mathrm{x}^2+\mathrm{y}^2+\frac{63}{12} \mathrm{x}+\frac{1}{2} \mathrm{y}=0 \\ & \therefore \text { centre }=\left(\frac{-63}{24}, \frac{-1}{4}\right) \end{aligned} $$ Now, $\mathrm{k}-2 \mathrm{~h}=\frac{-1}{4}+\frac{63}{12}=\frac{-3+63}{12}=5$

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

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