Given that we need to find the equation of the circle with the diagonal of a square as diameter.

It is also told that x = 6, x = 9, y = 3, and y = 6 are the sides of a square.

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

⇒ A = (6,3)

⇒ B = (9,3)

⇒ C = (9,6)

⇒ D = (6,6)

Since the diagonal of the square is the diameter of the circle, the circle circumscribes the square.

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

Let us assume diagonal BD as the diameter.

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

⇒ Centre(O) =

⇒

We have a circle with centre and passing through the point (6,3).

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 (x₁,y₁) and (x₂,y₂) 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 = r²

Now we substitute the corresponding values in the equation:

⇒

⇒

⇒ x² + y² - 15x - 9y + 72 = 0

∴The equation of the circle is x² + y² - 15x - 9y + 72 = 0.