TS EAMCET · Maths · Circle
The line \(x+2 y-c=0\) meets the curve \(x^2+y^2-3 x-6 y+3=0\) at two points \(\mathrm{P}\) and \(\mathrm{Q}\) and \(\angle \mathrm{POQ}=\frac{\pi}{2}\), where \(\mathrm{O}\) is the origin. Then \(2 c^2-15 \mathrm{c}=\)
- A 15
- B -15
- C 2
- D -2
Answer & Solution
Correct Answer
(B) -15
Step-by-step Solution
Detailed explanation
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