Let r be the radius of the circle, which touches the x-axis at point $$(a, 0)$$, $$a < 0$$ and the parabola $$y^2 = 9x$$ at the point $$(4, 6)$$. Then r is equal to _____.

The circle is tangent to the x-axis at the point $$\left(a,0\right)$$ with $$a \lt 0$$.
If a circle touches the x-axis, its centre must be exactly $$r$$ units above the point of contact.
Hence the centre of the required circle is $$C\left(a,r\right)$$ and its radius is $$r$$.

The circle is also tangent to the parabola $$y^{2}=9x$$ at the point $$P(4,6)$$.
Therefore two conditions hold:

1. CP = r   (distance condition)

2. CP is perpendicular to the tangent to the parabola at P   (normal condition)

Step 1: Distance condition

The distance CP is

$$CP=\sqrt{(4-a)^{2}+(6-r)^{2}}$$

Since CP = r, we get

$$(4-a)^{2}+(6-r)^{2}=r^{2} \qquad -(1)$$

Step 2: Normal condition

For the parabola $$y^{2}=9x$$, differentiate:
$$2y\,\frac{dy}{dx}=9 \;\;\Rightarrow\;\; \frac{dy}{dx}=\frac{9}{2y}$$

At P(4,6):
$$m_{\text{tangent}} = \frac{9}{2\cdot6} = \frac{3}{4}$$

Hence the slope of the normal is the negative reciprocal:

$$m_{\text{normal}} = -\frac{4}{3}$$

Equation of the normal through P:

$$y-6 = -\frac{4}{3}\,(x-4)$$

Since C(a,r) lies on this normal, substitute:

$$r-6 = -\frac{4}{3}\,(a-4)$$

Solve for r:

$$r = 6 - \frac{4}{3}(a-4)=6-\frac{4a}{3}+\frac{16}{3} = \frac{18}{3}+\frac{16}{3}-\frac{4a}{3} = \frac{34-4a}{3} \qquad -(2)$$

Step 3: Substitute into the distance equation

First compute $$6-r$$ using (2):
$$6-r = 6-\left(\frac{34-4a}{3}\right) = \frac{18}{3}-\frac{34-4a}{3} = \frac{18-34+4a}{3} = \frac{4a-16}{3} = \frac{4(a-4)}{3}$$

Put this in equation (1):

$$(4-a)^{2} + \left(\frac{4(a-4)}{3}\right)^{2} = \left(\frac{34-4a}{3}\right)^{2}$$

Multiply by 9 to clear denominators:

$$9(4-a)^{2} + 16(a-4)^{2} = (4a-34)^{2}$$

Because $$(a-4)^{2}=(4-a)^{2}$$, set $$s=(4-a)^{2}$$. Then

$$9s + 16s = 25s = (4a-34)^{2}$$

Step 4: Solve for a

Take square roots (all quantities are non-negative):

$$5|4-a| = |4a-34|$$

Since $$a \lt 0$$, $$4-a \gt 0$$ and $$4a-34 \lt 0$$, so

$$5(4-a) = -(4a-34) = 34-4a$$

$$20 - 5a = 34 - 4a$$

$$-a = 14 \quad\Longrightarrow\quad a = -14$$

Step 5: Find r

Insert $$a=-14$$ into (2):

$$r = \frac{34 - 4(-14)}{3} = \frac{34 + 56}{3} = \frac{90}{3} = 30$$

Hence the required radius is $$\mathbf{30}$$.