Q90 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics

If $C(\alpha, \beta)(\alpha < 0)$ is the centre of the circle that touches the Y-axis at $(0,3)$ and makes an intercept of length 2 units on positive $\mathrm{X}$-axis, then $(\alpha, \beta)=$

  • A
    $(-3, \sqrt{10})$
  • B
    $(-3, -\sqrt{10})$
  • Correct answer
    C
    $(-\sqrt{10}, 3)$
  • D
    $(-\sqrt{10}, -3)$

Solution

$2 \sqrt{g^2-c}=2 \Rightarrow g^2-c=1$ $\begin{aligned} & 2 \sqrt{f^2-c}=0 \Rightarrow f^2-c=0 \Rightarrow f^2=c \\ & x^2+y^2+2 g x+2 f y+c=0 \\ & \because \quad(0,3) \text { lies on circle } \\ & \quad 9+6 f+c=0 \\ & \Rightarrow f^2+6 f+9=0 \\ & \Rightarrow \quad(f+3)^2=0 \\ & \Rightarrow f=-3 \\ & \therefore \quad g^2-f^2=1 \\ & \Rightarrow \quad g^2=1+f^2=1+9 \Rightarrow g= \pm \sqrt{10} . \\ & \because \quad \text { Centre }(\alpha, \beta) \text { have } \alpha < 0 \\ & \quad \text { i.e. }-g < 0 \\ & \therefore \quad g=\sqrt{10} \\ & \therefore \quad(\alpha, \beta) \equiv(-\sqrt{10}, 3) . \end{aligned}$

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

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