Q88 / 150 · MHT CET 2024 (11 May Shift 1) · Mathematics
The equation of the tangent to the circle, given by $x=5 \cos \theta, y=5 \sin \theta$ at the point $\theta=\frac{\pi}{3}$ on it, is
- A$x-\sqrt{3} y=-5$
- Correct answerB$x+\sqrt{3} y=10$
- C$\sqrt{3} x+y=5 \sqrt{3}$
- D$\sqrt{3} x-y=0$
Solution
The equation of the tangent to the circle $x^2+y^2=\mathrm{a}^2$ at $\mathrm{P}(\theta)$ is $x \cos \theta+y \sin \theta=\mathrm{a}$
Here, $a=5, \theta=\frac{\pi}{3}$ $\therefore \quad$ The equation of the tangent is $\begin{aligned} & x \cos \frac{\pi}{3}+y \sin \frac{\pi}{3}=5 \\ & \Rightarrow x\left(\frac{1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=5 \\ & \Rightarrow x+y \sqrt{3}=10 \end{aligned}$
Here, $a=5, \theta=\frac{\pi}{3}$ $\therefore \quad$ The equation of the tangent is $\begin{aligned} & x \cos \frac{\pi}{3}+y \sin \frac{\pi}{3}=5 \\ & \Rightarrow x\left(\frac{1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=5 \\ & \Rightarrow x+y \sqrt{3}=10 \end{aligned}$
Hence, the correct answer is option: 'B'.