TS EAMCET · Maths · Circle
A tangent PT is drawn to the circle \(x^2+y^2=4\) at the point \(\mathrm{P}(\sqrt{3}, 1)\). If a straight line \(\mathrm{L}\) which is perpendicular to \(\mathrm{PT}\) is a tangent to the circle \((\mathrm{x}-3)^2+\mathrm{y}^2=1\), then a possible equation of \(\mathrm{L}\) is
- A \(x-\sqrt{3} y=1\)
- B \(x-\sqrt{3} y=4\)
- C \(x-\sqrt{3} y=-1\)
- D \(x-\sqrt{3} y=7\)
Answer & Solution
Correct Answer
(A) \(x-\sqrt{3} y=1\)
Step-by-step Solution
Detailed explanation
Equation of tangent to circle \(\mathrm{T}=0\) \(\Rightarrow\) T. \(\sqrt{3} x+y=4\)...(i) \[ m_{\mathrm{T}}=-\sqrt{3} \] \(\therefore\) Slope of line perpendicular to \(\mathrm{PT}=\frac{1}{\sqrt{3}}\) \(\because \mathrm{L}\) is tangent of \((x-3)^2+y^2=1\) i.e.…
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