AP EAMCET · Maths · Circle
Let \(\mathrm{A}(5,4)\) and \(\mathrm{B}(5,-4)\) be two points. If P is a point in the coordinate plane such that \(\underline{\mathrm{APB}}=\frac{\pi}{4}\), then the point P lies on the curve
- A \(x^2+y^2+10 x-17=0\)
- B \(x^2+y^2-2 x-31=0\)
- C \(x^2+y^2-10 x+17=0\)
- D \(x^2+y^2+2 x-31=0\)
Answer & Solution
Correct Answer
(B) \(x^2+y^2-2 x-31=0\)
Step-by-step Solution
Detailed explanation
\(m_{PA} = \frac{y-4}{x-5}\) \(m_{PB} = \frac{y+4}{x-5}\) \(\tan(\frac{\pi}{4}) = \left| \frac{m_{PB} - m_{PA}}{1 + m_{PA} m_{PB}} \right|\) \(1 = \left| \frac{\frac{y+4}{x-5} - \frac{y-4}{x-5}}{1 + \frac{(y-4)(y+4)}{(x-5)^2}} \right|\)…
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