Given that the circle has the radius 10 and has diameters 2x + y = 6 and 3x + 2y = 4.

We know that the centre is the intersection point of the diameters.

On solving the diameters, we get the centre to be (8, - 10).

We have a circle with centre (8, - 10) and having radius 10.

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:

⇒ (x - 8)² + (y - ( - 10))² = 10²

⇒ (x - 8)² + (y + 10)² = 100

⇒ x² - 16x + 64 + y² + 20y + 100 = 100

⇒ x² + y² - 16x + 20y + 64 = 0.

∴The equation of the circle is x² + y² - 16x + 20y + 64 = 0.