MHT CET · Maths · Circle
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 \(x-\sqrt{3} y=-5\)
- B \(x+\sqrt{3} y=10\)
- C \(\sqrt{3} x+y=5 \sqrt{3}\)
- D \(\sqrt{3} x-y=0\)
Answer & Solution
Correct Answer
(B) \(x+\sqrt{3} y=10\)
Step-by-step Solution
Detailed explanation
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
\(\begin{aligned}
& x \cos \frac{\pi}{3}+y \sin \frac{\pi}{3}=5 \\
& \Rightarrow x\left(\frac{1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=5 \\
& \Rightarrow x+y \sqrt{3}=10
\end{aligned}\)
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