JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Two tangents are drawn from a point \(P\) to the circle \(x^{2}+y^{2}-2 x-4 y+4=0\), such that the angle between these tangents is \(\tan ^{-1}\left(\frac{12}{5}\right)\), where \(\tan ^{-1}\left(\frac{12}{5}\right) \in(0, \pi)\). If the centre of the circle is denoted by \(C\) and these tangents touch the circle at points \(A\) and \(B\), then the ratio of the areas of \(\Delta PAB\) and \(\Delta CAB\) is :
- A \(11:4\)
- B \(9:4\)
- C \(3:1\)
- D \(2:1\)
Answer & Solution
Correct Answer
(B) \(9:4\)
Step-by-step Solution
Detailed explanation
\(\tan \theta=\frac{12}{5}\) \(PA =\cot \frac{\theta}{2}\) \(\therefore\) area of \(\Delta PAB =\frac{1}{2}( PA )^{2} \sin \theta=\frac{1}{2} \cot ^{2} \frac{\theta}{2} \sin \theta\) \(=\frac{1}{2}\left(\frac{1+\cos \theta}{1-\cos \theta}\right) \sin \theta\)…
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