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.