Question

Find the equation of the circle circumscribing the rectangle whose sides are x − 3y = 4, 3x + y = 22, x − 3y = 14 and 3x + y = 62.

Answer 1

Given:
Sides of the rectangle:
x − 3y = 4 ...(1)
3x + y = 22 ...(2)
x − 3y = 14 ...(3)
And, 3x + y = 62 ...(4)

The intersection of (1) and (2) is (7, 1).
The intersection of (2) and (3) is (8, −2).
The intersection of (3) and (4) is (20, 2).
The intersection of (1) and (4) is (19, 5).

Hence, the vertices of the rectangle are (7, −1), (8, −2), (20, 2) and (19, 5).

The vertices of the diagonals are (7, −1), (20, 2) and (19, 5), (8, −2).

Thus, the required equation of the circle is $\left(x-7\right)\left(x-20\right)+\left(y-1\right)\left(y-2\right)=0$ or $x^2+y^2-27x-3y+142=0$.