Question -
Answer -
Given:
The sides of a squaresare x = 6, x = 9, y = 3 and y = 6.
Let us assume A, B, C,D be the vertices of the square. On solving the lines, we get the coordinatesas: A = (6, 3)
B = (9, 3)
C = (9, 6)
D = (6, 6)
We know that theequation of the circle with diagonal AC is given by
(x – x1) (x– x2) + (y – y1) (y – y2) = 0
(x – 6) (x – 9) + (4 –3) (4 – 6) = 0
Upon simplifying, weget
x2 –6x – 9x + 54 + y2 – 3y – 6y + 18 = 0
x2 + y2 –15x – 9y + 72 = 0
We know that theequation of the circle with diagonal BD as diameter is given by
(x – x1) (x– x2) + (y – y1) (y – y2) = 0
(x – 9) (x – 6) + (y –3) (y – 6) = 0
Upon simplifying, weget
x2 –9x – 6x + 54 + y2 – 3y – 6y + 18 = 0
x2 + y2 –15x – 9y + 72 = 0
∴The equation of thecircle is x2 + y2 – 15x – 9y + 72 = 0