AP EAMCET · Maths · Circle
Let \(P Q\) and \(R S\) be tangents at the extremities of a diameter \(P R\) of a circle of radius \(r\) such that \(P S\) and \(R Q\) intersect at a point \(X\) on the circumference of the circle, then \(2 r\) equals
- A \(\sqrt{P Q \cdot R S}\)
- B \(\frac{P Q+R S}{2}\)
- C \(\frac{2 P Q \cdot R S}{P Q+R S}\)
- D \(\sqrt{\frac{(P Q)^2+(R S)^2}{2}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{P Q \cdot R S}\)
Step-by-step Solution
Detailed explanation
According to the question, from the diagram, in \(\triangle P Q R\) \(\tan \theta=\frac{P Q}{P R} \Rightarrow P R=P Q \cot \theta\)...(i) and in \(\triangle P R S\),…
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