Question

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 PA² + PB² = AB².

Answer 1

To find relation between x and y, if PA² + PB² = AB²

`PA = sqrt((x - 3)^2 + (y - 0)^2)`   ...(∴ Distance formula)

PA² = (x – 3)² + (y)²   ...(∴ Squaring on both sides)

`PB = sqrt((x - 0)^2 + (y - 4)^2)`   ...(∴ Distance formula)

PB² = (x)² + (y – 4)²    ...(∴ Squaring on both sides)

`AB = sqrt((3 - 0)^2 + (0 - 4)^2)`   ...(∴ Distance formula)

AB² = (3)² + (–4)²

AB² = 9 + 16

AB² = 25

Now, PA² + PB² = AB²

(x – 3)² + y² + x² + (y – 4)² = 25

x² – 6x + 9 + y² + x² + y² – 8y + 16 = 25   ...(∴ (a – b)² = a² + b² – 2ab)

2x² + 2y² – 6x – 8y + 25 = 25

2x² + 2y² – 6x – 8y = 25 – 25 = 0

2x² + 2y² – 6x – 8y = 0

2[x² + y² – 3x – 4y] = 0

x² + y² – 3x – 4y = 0 which is required relation.