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

Since, the centre is the intersection point of the diameters.

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

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

By using the formula,

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

Where, p = 8, q = -10, r = 10

substituting the values in the equation, we get

\[\begin{array}{*{35}{l}}

{{\left( x\text{ }-\text{ }8 \right)}^{2}}~+\text{ }{{\left( y\text{ }-\text{ }\left( -10 \right) \right)}^{2}}~=\text{ }{{10}^{2}}  \\

{{\left( x\text{ }-\text{ }8 \right)}^{2}}~+\text{ }{{\left( y\text{ }+\text{ }10 \right)}^{2}}~=\text{ }100  \\

{{x}^{2}}~-\text{ }16x\text{ }+\text{ }64\text{ }+\text{ }{{y}^{2}}~+\text{ }20y\text{ }+\text{ }100\text{ }=\text{ }100  \\

{{x}^{2}}~+\text{ }{{y}^{2}}~-\text{ }16x\text{ }+\text{ }20y\text{ }+\text{ }64\text{ }=\text{ }0.  \\

\end{array}\]

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