Question:

**Step 1: Set up coordinates for the circles and common tangent.** Let the common external tangent be the x-axis, $y=0$. Let the circle with radius $r_1=36$ have its center $O_1$ on the y-axis. Thus, $O_1=(0, 36)$. Let the circle with radius $r_2=9$ have its center $O_2=(x_2, 9)$. Since the two circles touch externally, the distance between their centers is $r_1+r_2 = 36+9=45$. Using the distance formula for $O_1O_2$: $O_1O_2^2 = (x_2-0)^2 + (9-36)^2 = 45^2$ $x_2^2 + (-27)^2 = 45^2$ $x_2^2 + 729 = 2025$ $x_2^2 = 1296$ $x_2 = 36$ (We choose the positive x-coordinate for $O_2$ without loss of generality). Thus, the centers of the two given circles are $O_1=(0, 36)$ and $O_2=(36, 9)$. **Step 2: Set up conditions for the unknown circle.** Let the unknown circle have center $P=(x,y)$ and radius $r$. Since this circle touches the common tangent $y=0$ externally, its y-coordinate must be $r$. So, $P=(x,r)$. **Step 3: Apply external tangency to the first circle.** The distance from the center of the unknown circle $P=(x,r)$ to the center of the first circle $O_1=(0,36)$ is $r+r_1 = r+36$. $\sqrt{(x-0)^2 + (r-36)^2} = r+36$ Squaring both sides: $x^2 + (r-36)^2 = (r+36)^2$ $x^2 + r^2 - 72r + 1296 = r^2 + 72r + 1296$ $x^2 = 144r$ **Step 4: Apply external tangency to the second circle.** The distance from the center of the unknown circle $P=(x,r)$ to the center of the second circle $O_2=(36,9)$ is $r+r_2 = r+9$. $\sqrt{(x-36)^2 + (r-9)^2} = r+9$ Squaring both sides: $(x-36)^2 + (r-9)^2 = (r+9)^2$ $(x-36)^2 + r^2 - 18r + 81 = r^2 + 18r + 81$ $(x-36)^2 = 36r$ **Step 5: Solve for $r$.** From Step 3, $x = \pm 12\sqrt{r}$. We assume the unknown circle is located between the two given circles, so $0 < x < 36$, which implies $x=12\sqrt{r}$. Substitute $x=12\sqrt{r}$ into the equation from Step 4: $(12\sqrt{r} - 36)^2 = 36r$ Divide both sides by 36: $\left(\frac{12\sqrt{r} - 36}{6}\right)^2 = r$ $(2\sqrt{r} - 6)^2 = r$ Expand the left side: $4r - 24\sqrt{r} + 36 = r$ Rearrange the terms to form a quadratic equation in $\sqrt{r}$: $3r - 24\sqrt{r} + 36 = 0$ Divide by 3: $r - 8\sqrt{r} + 12 = 0$ Let $u = \sqrt{r}$. The equation becomes: $u^2 - 8u + 12 = 0$ Factor the quadratic equation: $(u-2)(u-6) = 0$ This yields two possible values for $u$: $u=2$ or $u=6$. Since $u=\sqrt{r}$: If $u=2$, then $\sqrt{r}=2 \implies r=4$. If $u=6$, then $\sqrt{r}=6 \implies r=36$. The problem asks for "The radius of the circle", implying a unique answer. The smaller radius, corresponding to the circle nestled between the two larger ones, is typically the intended solution. The radius of the circle is 4.