Answer :

Given that we need to find the equation of the circle which circumscribes the rectangle.

It is also told that x - 3y = 4, 3x + y = 22, x - 3y = 14 and 3x + y = 62 are the sides of a rectangle.

Let us assume A,B,C,D are the vertices of the rectangle. On solving the lines, we get the vertices as:

⇒ A = (7,1)

⇒ B = (8, - 2)

⇒ C = (20,2)

⇒ D = (19,5)

Since the circle circumscribes the rectangle, the diagonal of the rectangle will be the diameter of the circle.

So, taking any diagonal as diameter gives the same equation of the circle.

Let us assume diagonal AC as the diameter.

We know that the centre is the mid - point of the diameter.

⇒ Centre(C) =

⇒

We have a circle with a centre and passing through the point (7,1).

We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.

We know that the distance between the two points (x_{1},y_{1}) and (x_{2},y_{2}) is .

Let us assume r is the radius of the circle.

⇒

⇒

⇒

⇒

⇒

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:

⇒ (x - p)^{2} + (y - q)^{2} = r^{2}

Now we substitute the corresponding values in the equation:

⇒

⇒

⇒ x^{2} + y^{2} - 27x - 3y + 142 = 0

∴The equation of the circle is x^{2} + y^{2} - 27x - 3y + 142 = 0.

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