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(A) The equation of the circle touching the $x$-axis at $(a, 0)$ with radius $r$ is $(x - a)^2 + (y - r)^2 = r^2$.Since it passes through $(4, 6)$,we

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$30$

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(A) The equation of the circle touching the $x$-axis at $(a, 0)$ with radius $r$ is $(x - a)^2 + (y - r)^2 = r^2$.Since it passes through $(4, 6)$,we have $(4 - a)^2 + (6 - r)^2 = r^2$.Expanding this,$16 - 8a + a^2 + 36 - 12r + r^2 = r^2$,which simplifies to $a^2 - 8a - 12r + 52 = 0$ (Equation $1$).The tangent to the parabola $y^2 = 9x$ at $(4, 6)$ is given by $y(6) = \frac{9}{2}(x + 4)$,which simplifies to $12y = 9x + 36$,or $3x - 4y + 12 = 0$.Since the circle is tangent to this line at $(4, 6)$,the distance from the center $(a, r)$ to the line $3x - 4y + 12 = 0$ must be equal to $r$.Thus,$\frac{|3a - 4r + 12|}{\sqrt{3^2 + (-4)^2}} = r$,which gives $|3a - 4r + 12| = 5r$.This implies $3a - 4r + 12 = 5r$ or $3a - 4r + 12 = -5r$.Case $1$: $3a - 9r + 12 = 0 \Rightarrow a = 3r - 4$. Substituting into Equation $1$: $(3r - 4)^2 - 8(3r - 4) - 12r + 52 = 0$.$9r^2 - 24r + 16 - 24r + 32 - 12r + 52 = 0$ $\Rightarrow 9r^2 - 60r + 100 = 0$ $\Rightarrow (3r - 10)^2 = 0$ $\Rightarrow r = \frac{10}{3}$.Case $2$: $3a + r + 12 = 0 \Rightarrow a = \frac{-r - 12}{3}$. Substituting into Equation $1$: $(\frac{-r - 12}{3})^2 - 8(\frac{-r - 12}{3}) - 12r + 52 = 0$.Multiplying by $9$: $(r + 12)^2 + 24(r + 12) - 108r + 468 = 0$.$r^2 + 24r + 144 + 24r + 288 - 108r + 468 = 0$ $\Rightarrow r^2 - 60r + 900 = 0$ $\Rightarrow (r - 30)^2 = 0$ $\Rightarrow r = 30$.Since $a < 0$,for $r = 30$,$a = \frac{-30 - 12}{3} = -14 < 0$,which is valid. Thus,$r = 30$.

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

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