Question -
Answer -
Given:
The sides x – 3y = 4…. (1)
3x + y = 22 … (2)
x – 3y = 14 …. (3)
3x + y = 62 … (4)
Let us assume A, B, C,D be the vertices of the square. On solving the lines, we get the coordinatesas: A = (7, 1)
B = (8, – 2)
C = (20, 2)
D = (19, 5)
We know that theequation of the circle with diagonal AC as diameter is given by
(x – x1) (x– x2) + (y – y1) (y – y2) = 0
(x – 7) (x – 20) + (y– 1) (y – 2) = 0
Upon simplification weget
x2 + y2 –27x – 3y + 142 = 0
∴The equation of thecircle is x2 + y2 – 27x – 3y + 142 = 0