JEE Advanced · Mathematics · 12. Circle
A tangent \(P T\) is drawn to the circle \(x^{2}+y^{2}=4\) at the point \(P(\sqrt{3}, 1)\). A straight line \(L\), perpendicular to \(P T\) is a tangent to the circle \((x-3)^{2}+y^{2}-1\).
Question: A possible equation of \(L\) is
- A \(x-\sqrt{3} y=1\)
- B \(x+\sqrt{3} y=1\)
- C \(x-\sqrt{3} y=-1\)
- D \(x+\sqrt{3} y=5\)
Answer & Solution
Correct Answer
(A) \(x-\sqrt{3} y=1\)
Step-by-step Solution
Detailed explanation
Equation of tangent \(P T\) to the circle \(x^{2}+y^{2}=4\) at the point \(P(\sqrt{3}, 1)\) is \(x \sqrt{3}+y=4\) Let the line \(L\), perpendicular to tangent \(P T\) be \(x-y \sqrt{3}+\lambda=0\)
As it is tangent to the circle \((x-3)^{2}+y^{2}=1\) \(\therefore \quad\) Length of perpendicular from centre of circle to the Tangent \(=\) radius of circle.
\(\Rightarrow\left|\frac{3+\lambda}{2}\right|=1 \Rightarrow \lambda=-1 \text { or }-5\)
\(\therefore\) Equation of \(L\) can be \(x-\sqrt{3} y=1\) or \(x-\sqrt{3} y=5\)
As it is tangent to the circle \((x-3)^{2}+y^{2}=1\) \(\therefore \quad\) Length of perpendicular from centre of circle to the Tangent \(=\) radius of circle.
\(\Rightarrow\left|\frac{3+\lambda}{2}\right|=1 \Rightarrow \lambda=-1 \text { or }-5\)
\(\therefore\) Equation of \(L\) can be \(x-\sqrt{3} y=1\) or \(x-\sqrt{3} y=5\)
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