Convert the equation 1/2x2 + 1/2y2 = 3x − 5y − 9 to standard form and find the circle's radius and center.

The given equation is:

$$\frac{1}{2}x^2 + \frac{1}{2}y^2 = 3x - 5y - 9$$

Move the $x$ and $y$ terms to the left side:

$$\frac{1}{2}x^2 - 3x + \frac{1}{2}y^2 + 5y = -9$$

$$2 \left( \frac{1}{2}x^2 - 3x + \frac{1}{2}y^2 + 5y \right) = 2(-9)$$ $$x^2 - 6x + y^2 + 10y = -18$$

To complete the square for $x^2 - 6x$, take half of the coefficient of $x$ (which is $-6$), square it ($(-3)^2 = 9$), and add it to both sides.
To complete the square for $y^2 + 10y$, take half of the coefficient of $y$ (which is 10), square it ($(5)^2 = 25$), and add it to both sides.

$$(x^2 - 6x + 9) + (y^2 + 10y + 25) = -18 + 9 + 25$$

$$(x - 3)^2 + (y + 5)^2 = -18 + 34$$ $$(x - 3)^2 + (y + 5)^2 = 16$$

Comparing $(x - 3)^2 + (y + 5)^2 = 16$ with $(x-h)^2 + (y-k)^2 = r^2$:
The center $(h, k)$ is $(3, -5)$.
The radius squared $r^2$ is 16, so the radius $r = \sqrt{16} = 4$.

The coordinates of the center of the circle are $\boxed{(3, -5)}$ and the radius is $\boxed{4}$.

I cannot draw the circle in the grid, but you would plot the center at $(3, -5)$ and then draw a circle with a radius of 4 units around that center.