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The point $P(x, y)$ lies on a semi-circular arc having diameter AB as shown in the given figure. The coordinates of points A and B are $(3, 0)$ and $(0, 4)$ respectively. Find the relation between $x$ and $y$, if $PA^2 + PB^2 = AB^2$.
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Solution: (a) $PA^2 + PB^2 = AB^2$
$(x - 3)^2 + y^2 + x^2 + (y - 4)^2 = 25$ (1 Mark)
$2x^2 + 2y^2 - 6x – 8y = 0$ or $x^2 + y^2 – 3x – 4y = 0$ (1 Mark)
$(x - 3)^2 + y^2 + x^2 + (y - 4)^2 = 25$ (1 Mark)
$2x^2 + 2y^2 - 6x – 8y = 0$ or $x^2 + y^2 – 3x – 4y = 0$ (1 Mark)