The circle passing through the intersection of the circles, $$x^2 + y^2 - 6x = 0$$ and $$x^2 + y^2 - 4y = 0$$ having its centre on the line, $$2x - 3y + 12 = 0$$, also passes through the point:

The two given circles are

$$x^{2}+y^{2}-6x = 0 \qquad\text{and}\qquad x^{2}+y^{2}-4y = 0.$$

The standard method to obtain every circle that passes through the common points (their “radical family”) is to write

$$\bigl(x^{2}+y^{2}-6x\bigr) \;+\;\lambda\,\bigl(x^{2}+y^{2}-4y\bigr)=0,$$

where $$\lambda$$ is a real parameter. Expanding the brackets we get

$$\;(1+\lambda)\,(x^{2}+y^{2})\;-\;6x\;-\;4\lambda\,y\;=\;0.$$

To identify the centre easily, we divide every term by the common factor $$1+\lambda$$ (allowed as long as $$\lambda\neq-1$$, which will turn out to be true):

$$x^{2}+y^{2}\;-\;\dfrac{6}{1+\lambda}\,x\;-\;\dfrac{4\lambda}{1+\lambda}\,y = 0.$$

A circle in general form is written as

$$x^{2}+y^{2}-2gx-2fy+c=0,$$

whose centre is $$\bigl(g,f\bigr).$$ Comparing coefficients, we obtain

$$2g=\dfrac{6}{1+\lambda}\;\Longrightarrow\;g=\dfrac{3}{1+\lambda},$$

$$2f=\dfrac{4\lambda}{1+\lambda}\;\Longrightarrow\;f=\dfrac{2\lambda}{1+\lambda}.$$

Therefore the centre of the required circle is

$$\left(\dfrac{3}{1+\lambda},\;\dfrac{2\lambda}{1+\lambda}\right).$$

It is given that this centre lies on the straight line

$$2x-3y+12=0.$$

Substituting $$x=\dfrac{3}{1+\lambda}$$ and $$y=\dfrac{2\lambda}{1+\lambda}$$ into the line’s equation, we have

$$2\left(\dfrac{3}{1+\lambda}\right)-3\left(\dfrac{2\lambda}{1+\lambda}\right)+12=0.$$

The denominators are the same, so we combine the fractions:

$$\dfrac{6-6\lambda}{1+\lambda}+12=0.$$

To clear the denominator, multiply every term by $$1+\lambda$$:

$$6-6\lambda+12(1+\lambda)=0.$$

Expanding the last bracket,

$$6-6\lambda+12+12\lambda = 0.$$

Collecting like terms,

$$(6+12)+( -6\lambda+12\lambda)=0 \;\;\Longrightarrow\;\; 18+6\lambda=0.$$

Dividing by $$6$$ we find

$$\lambda=-3.$$

Now we substitute $$\lambda=-3$$ back into the family to get the explicit equation of the desired circle. First compute $$1+\lambda=1-3=-2.$$ Then

$$(1+\lambda)(x^{2}+y^{2})-6x-4\lambda y=0$$ $$\Longrightarrow\;(-2)(x^{2}+y^{2})-6x-4(-3)y=0$$ $$\Longrightarrow\;-2x^{2}-2y^{2}-6x+12y=0.$$

Multiplying by $$-1$$ to simplify,

$$2x^{2}+2y^{2}+6x-12y=0,$$

and dividing by $$2$$ gives

$$x^{2}+y^{2}+3x-6y=0.$$

Any point lying on this circle must satisfy the equation above. We now test the four options.

Option A: $$(-1,3)$$ $$(-1)^{2}+3^{2}+3(-1)-6(3)=1+9-3-18 = -11\neq0.$$

Option B: $$(-3,6)$$ $$(-3)^{2}+6^{2}+3(-3)-6(6)=9+36-9-36 = 0.$$

Option C: $$(-3,1)$$ $$(-3)^{2}+1^{2}+3(-3)-6(1)=9+1-9-6 = -5\neq0.$$

Option D: $$(1,-3)$$ $$1^{2}+(-3)^{2}+3(1)-6(-3)=1+9+3+18 = 31\neq0.$$

Only the point $$(-3,6)$$ satisfies the circle’s equation.

Hence, the correct answer is Option B.