Let A(5,4) and B(5,-4) be two points. If P(x, | vedclass

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(B) Let $P = (x, y)$,$A = (5, 4)$,and $B = (5, -4)$.Given $	an(\angle APB) = 	an(\frac{\pi}{4}) = 1$.The slope of $PA$ is $m_1 = \frac{y-4}{x-5}$ an

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$x^2+y^2-10x+2y+17=0$

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(B) Let $P = (x, y)$,$A = (5, 4)$,and $B = (5, -4)$.Given $	an(\angle APB) = 	an(\frac{\pi}{4}) = 1$.The slope of $PA$ is $m_1 = \frac{y-4}{x-5}$ and the slope of $PB$ is $m_2 = \frac{y+4}{x-5}$.Using the formula $	an 	heta = |\frac{m_1-m_2}{1+m_1m_2}| = 1$,we have:$|\frac{\frac{y-4}{x-5} - \frac{y+4}{x-5}}{1 + \frac{(y-4)(y+4)}{(x-5)^2}}| = 1$$|\frac{-8/(x-5)}{(x-5)^2 + y^2 - 16 / (x-5)^2}| = 1$$|\frac{-8(x-5)}{(x-5)^2 + y^2 - 16}| = 1$$| -8(x-5) | = | (x-5)^2 + y^2 - 16 |$$| -8x + 40 | = | x^2 - 10x + 25 + y^2 - 16 |$$| -8x + 40 | = | x^2 + y^2 - 10x + 9 |$This implies $x^2 + y^2 - 10x + 9 = \pm(-8x + 40)$.Case $1$: $x^2 + y^2 - 10x + 9 = -8x + 40 \implies x^2 + y^2 - 2x - 31 = 0$.Case $2$: $x^2 + y^2 - 10x + 9 = 8x - 40 \implies x^2 + y^2 - 18x + 49 = 0$.Comparing with the options,the correct curve is $x^2+y^2-2x-31=0$.

Let $A(5,4)$ and $B(5,-4)$ be two points. If $P(x,y)$ is a point in the coordinate plane such that $\angle APB = \frac{\pi}{4}$,then the point $P$ lies on the curve

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