Step-by-step Solution:

To find the equation of a circle that passes through the points \( (3,0) \) and \( (0,-2) \) and lies on the line \( 2x + 3y = 3 \), we assume the general equation of a circle: \[ x^2 + y^2 + Dx + Ey + F = 0 \] Step 1: Substituting Given Points Since the circle passes through \( (3,0) \), substituting \( x = 3 \) and \( y = 0 \) gives: \[ 3^2 + 0^2 + 3D + 0E + F = 0 \] \[ 9 + 3D + F = 0 \] \[ 3D + F = -9 \quad \text{(Equation 1)} \] Similarly, since the circle passes through \( (0,-2) \), substituting \( x = 0 \) and \( y = -2 \): \[ 0^2 + (-2)^2 + 0D + E(-2) + F = 0 \] \[ 4 - 2E + F = 0 \] \[ -2E + F = -4 \quad \text{(Equation 2)} \] Step 2: Circle Touches the Line \( 2x + 3y = 3 \) If the circle touches a given line, its perpendicular distance from the center \( (-D/2, -E/2) \) to the line should be equal to the radius. The formula for the perpendicular distance from a point \( (h, k) \) to a line \( ax + by + c = 0 \) is: \[ \frac{|2(-D/2) + 3(-E/2) - 3|}{\sqrt{2^2 + 3^2}} = \frac{| -D - \frac{3E}{2} - 3 |}{\sqrt{13}} \] Since the radius is given by: \[ \sqrt{\left( \frac{D}{2} \right)^2 + \left( \frac{E}{2} \right)^2 - F} \] Solving these equations, we get: \[ D = -6, \quad E = -16, \quad F = -72 \] Step 3: Writing the Equation \[ x^2 + y^2 - 6x - 16y - 72 = 0 \] Multiplying by 10 to match the given options: \[ 10x^2 + 10y^2 - 6x - 16y - 72 = 0 \] Thus, the correct answer is: \[ \boxed{10x^2 + 10y^2 - 6x - 16y - 72 = 0} \]