Q91 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics

The equations of the tangents to the circle $x^2+y^2=4$ drawn from the point $(4,0)$ are

  • Correct answer
    A
    $\sqrt{3} y= \pm(x-4)$
  • B
    $\sqrt{3} y= \pm2(x-4)$
  • C
    $\sqrt{3} x= \pm(y-4)$
  • D
    $\sqrt{3} x= \pm2(y-4)$

Solution

Equation of pair of tangents, $S S_1=T^2$ $\begin{aligned} & \Rightarrow \quad\left(x^2+y^2-4\right)(16+0-4)=(4 x-4)^2 \\ & \Rightarrow \quad 12\left(x^2+y^2-4\right)=16(x-1)^2 \\ & \Rightarrow \quad 3\left(x^2+y^2-4\right)=4\left(x^2+1-2 x\right) \\ & \Rightarrow x^2-3 y^2-8 x+16=0 \\ & \Rightarrow \quad(x-4)^2-3 y^2=0 \\ & \Rightarrow \quad 3 y^2=(x-4)^2 \\ & \Rightarrow \quad \sqrt{3} y= \pm(x-4)\end{aligned}$

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

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