If $$y + 3x = 0$$ is the equation of a chord of the circle $$x^2 + y^2 - 30x = 0$$, then the equation of the circle with this chord as diameter is:

We are told that the given circle is $$x^2 + y^2 - 30x = 0$$. First, we rewrite this in the standard form $$(x - h)^2 + (y - k)^2 = r^2$$ so that its centre and radius become clear.

Completing the square in the $$x$$-terms only (because no $$y$$-term is present):

$$x^2 - 30x = x^2 - 30x + 225 - 225 = (x - 15)^2 - 225.$$

Hence

$$x^2 + y^2 - 30x = (x - 15)^2 + y^2 - 225 = 0 \;\;\Longrightarrow\;\; (x - 15)^2 + y^2 = 225.$$

So the centre of the given circle is $$(15,\,0)$$ and its radius is $$15.$$

The chord whose equation is given is $$y + 3x = 0,$$ or equivalently $$y = -3x.$$ We next find the two points where this line meets the circle, because those points will be the end-points of the chord.

Substituting $$y = -3x$$ into the circle’s equation:

$$x^2 + (-3x)^2 - 30x = 0 \;\;\Longrightarrow\;\; x^2 + 9x^2 - 30x = 0 \;\;\Longrightarrow\;\; 10x^2 - 30x = 0.$$

Factoring out $$10x$$ gives

$$10x(x - 3) = 0 \;\;\Longrightarrow\;\; x = 0 \quad \text{or} \quad x = 3.$$

Putting these $$x$$-values back into $$y = -3x:$$

For $$x = 0: \; y = -3(0) = 0.$$ For $$x = 3: \; y = -3(3) = -9.$$

Thus the chord meets the circle at the two points $$P_1(0,\,0) \quad\text{and}\quad P_2(3,\,-9).$$

It is stated that this chord is to serve as the diameter of a new circle. For any circle, the centre is the midpoint of any diameter. Therefore, the centre of the required circle is the midpoint of $$P_1$$ and $$P_2.$$

The midpoint formula says $$\bigl(\tfrac{x_1 + x_2}{2},\,\tfrac{y_1 + y_2}{2}\bigr).$$ Applying it:

$$h = \frac{0 + 3}{2} = \frac{3}{2}, \qquad k = \frac{0 + (-9)}{2} = \frac{-9}{2}.$$

So the centre of the required circle is $$\bigl(\tfrac{3}{2},\, -\tfrac{9}{2}\bigr).$$

Next, we need the radius of the new circle. Because $$P_1P_2$$ is a diameter, the radius is half the length of $$P_1P_2.$$ The distance formula for two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$ Using $$P_1(0,0)$$ and $$P_2(3,-9):$$

$$\text{Diameter length} = \sqrt{(3 - 0)^2 + (-9 - 0)^2} = \sqrt{9 + 81} = \sqrt{90}.$$

Therefore

$$\text{Radius} = \frac{\sqrt{90}}{2}, \qquad \text{so } r^2 = \frac{90}{4} = \frac{45}{2}.$$

Now we write the equation of a circle with centre $$(h,\,k)$$ and radius squared $$r^2:$$ $$(x - h)^2 + (y - k)^2 = r^2.$$ Substituting $$h = \frac{3}{2},\; k = -\frac{9}{2},\; r^2 = \frac{45}{2}:$$

$$\bigl(x - \tfrac{3}{2}\bigr)^2 + \bigl(y + \tfrac{9}{2}\bigr)^2 = \frac{45}{2}.$$

We expand each square term completely:

$$\bigl(x - \tfrac{3}{2}\bigr)^2 = x^2 - 2\!\times\!x\!\times\!\tfrac{3}{2} + \bigl(\tfrac{3}{2}\bigr)^2 = x^2 - 3x + \frac{9}{4},$$

$$\bigl(y + \tfrac{9}{2}\bigr)^2 = y^2 + 2\!\times\!y\!\times\!\tfrac{9}{2} + \bigl(\tfrac{9}{2}\bigr)^2 = y^2 + 9y + \frac{81}{4}.$$

Adding the two expansions:

$$x^2 - 3x + \frac{9}{4} \;+\; y^2 + 9y + \frac{81}{4} = \frac{45}{2}.$$

Combine the constant fractions on the left:

$$\frac{9}{4} + \frac{81}{4} = \frac{90}{4} = \frac{45}{2}.$$

So we have

$$x^2 + y^2 - 3x + 9y + \frac{45}{2} = \frac{45}{2}.$$

Subtracting $$\frac{45}{2}$$ from both sides eliminates the constant term, leaving

$$x^2 + y^2 - 3x + 9y = 0.$$

This matches exactly with Option B in the list.

Hence, the correct answer is Option B.