JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let \(O\) be the origin and \(OP\) and \(OQ\) be the tangents to the circle \(x^2+y^2-6 x+4 y+8=0\) at the point \(P\) and \(Q\) on it. If the circumcircle of the triangle OPQ passes through the point \(\left(\alpha, \frac{1}{2}\right)\), then a value of \(\alpha\) is
- A \(\frac{3}{2}\)
- B \(\frac{5}{2}\)
- C \(1\)
- D \(-\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{5}{2}\)
Step-by-step Solution
Detailed explanation
Circumcircle of \(\triangle OPQ\) \((x-0)(x-3)+(y-0)(y+2)=0\) \(x^2+y^2-3 x+2 y=0\) \(\text { passes through }\left(\alpha, \frac{1}{2}\right)\) \(\therefore \alpha^2+\frac{1}{4}-3 \alpha+1=0\) \(\Rightarrow \alpha^2-3 \alpha+\frac{5}{4}=0 \Rightarrow 4 \alpha^2-12 \alpha+5=0\)…
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