Q91 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

$C_1$ and $C_2$ are the external and internal centres of similitude of the circles $x^2+y^2-2 x+4 y+1=0$ and $x^2+y^2+4 x-6 y+12=0$. If the radius of the circle having $C_1 C_2$ as its diameters is $r$, then $\frac{9}{2} r=$

  • A
    $\sqrt{15}$
  • B
    $3 \sqrt{15}$
  • C
    $2 \sqrt{34}$
  • Correct answer
    D
    $3 \sqrt{34}$

Solution

$C_1$ and $C_2$ are external and internal similitude of the circle $x^2+y^2-2 x+4 y+1=0$ and $\quad x^2+y^2+4 x-6 y+12=0$ Centre of circle $x^2+y^2-2 x+4 y+1=0$ is $(1,-2), r_1=2$ Centre of circle $x^2+y^2+4 x-6 y+4=0$ $(-2,3), r_2=1$ Coordinate of $C_1=\frac{-5}{3}$ $\left(\frac{2(-2)-1}{2-1}, \frac{6+2}{2-1}\right) \equiv(-5,8)$ Coordinate of $C_2=\left(\frac{2(-2)+1}{2+1}, \frac{6-2}{2+1}\right) \equiv\left(-1, \frac{4}{3}\right)$ $C_1 C_2=\sqrt{(-5+1)^2+\left(8-\frac{4}{3}\right)^2}$ $C_1 C_2=\sqrt{16+\frac{400}{9}} \Rightarrow C_1 C_2=\sqrt{\frac{544}{9}}=\frac{4 \sqrt{34}}{3}$ Here, $C_1 C_2=2 r$ $\therefore \quad \frac{4}{3} \sqrt{34}=2 r$ $\frac{9}{2} r=3 \sqrt{34}$

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

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