The equation of the tangent to the circle $x^2+y^2-9=0$ making an angle $60^{\circ}$ with the $X$-axis is

The equation of the tangent to the circle $x^2+y^2-9=0$ making an angle $60^{\circ}$ with the $X$-axis is
  1. $\frac{1}{\sqrt{3}} x-y \pm 6=0$
  2. $\sqrt{3} x-y \pm 6=0$
  3. $\sqrt{3} x+y \pm 6=0$
  4. $\frac{1}{\sqrt{3}} x+y \pm 6=0$

Solution

Slope of tangent $=m=\tan 60^{\circ}=\sqrt{3}$ Equation of tangent, $y=m x \pm a \sqrt{1+m^2}$ $\Rightarrow \quad y=\sqrt{3} x \pm 3 \sqrt{1+3} \quad[\because$ radius $=3]$ $\begin{array}{ll}\Rightarrow & y=\sqrt{3} x \pm 6 \\ \Rightarrow & \sqrt{3} x-y \pm 6=0\end{array}$

Asked in: AP EAMCET 2022 (07 Jul Shift 1)

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