TS EAMCET · Maths · Circle
From a point \(P\) on the circle \(x^2+y^2-4 x-6 y+9=0\), a pair of tangents \(P Q\) and \(P R\) are drawn touching the circle \(x^2+y^2-4 x-6 y+12=0\) at \(Q\) and \(R\). If \(C\) is the centre of the concentric circles, then the area of the \(\triangle C Q R\) (in sq. units) is
- A \(\frac{1}{2}\)
- B \(\frac{\sqrt{3}}{2}\)
- C \(\frac{\sqrt{3}}{4}\)
- D \(\frac{3}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sqrt{3}}{4}\)
Step-by-step Solution
Detailed explanation
Given equation of circle, \(\begin{array}{r} x^2+y^2-4 x-6 y+9=0 \\ R=P C=\sqrt{4+9-9}=2 \\ x^2+y^2-4 x-6 y+12=0 \\ r=Q C=\sqrt{4+9-12}=1 \end{array}\) In \(\triangle P Q C\), \(\cos \theta=\frac{Q C}{P C}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}\) \(\therefore\) Area of…
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