$A, B, C$ , and $D$ are four points in a straight line; prove that the locus of a point $P$ , such that the angles $A P B$ and $C P D$ are equal, is a circle.

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Solution

Let A be the origin
The line along ABCD be the axis of $^{'} x^{'}$ and the line perpendicular to it at A as y-axis.
$P (h, k)$ be any point.
Let the co ordinates of B, C and D be $(b, 0), (c, 0)$ and $(d, 0)$ respectively. A is $(0, 0)$
$∴$ if the slope of AP, BP, CP and DP be $m_{1}, m_{2}, m_{3}$ and $m_{4}$ respectively, then
$m_{1} = \frac{k}{h} . . . . .1 m_{2} = \frac{k}{h - b} . . . . .2$
$m_{3} = \frac{k}{h - c} . . . . . . .3 m_{4} = \frac{k}{h - d} . . . . . .4$
$∠ A P B = ∠ C P D$
$tan A P B = tan C P D$
Hence, $\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} = \frac{m_{3} - m_{4}}{1 + m_{3} m_{4}}$
or $\frac{k h - k b - k h}{h^{2} - h b + k^{2}} = \frac{k h - k d - k h + k c}{h^{2} - h d - h c + c d + k^{2}}$
or $b (h^{2} + k^{2} - h d - h c + c d) = (c - d) (h^{2} - b h + k^{2})$ as $k \neq 0$
or $h^{2} (b - c + d) + k^{2} (b - c + d) + h (b c - b d - b d - b c) + b c d = 0$
$x^{2} (b - c + d) + y^{2} (b - c + d) - 2 b d x + b c d = 0$
It is a circle

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A,B,C, and D are four points in a straight line; prove that the locus of a point P, such that the angles APB and CPD are equal, is a circle.

$A, B, C$ , and $D$ are four points in a straight line; prove that the locus of a point $P$ , such that the angles $A P B$ and $C P D$ are equal, is a circle.