Question:

Step 1: Set up the geometric configuration. Let $O$ be the common center of all three concentric circles. We are given: - Circle $C_1$ has radius $1$ - Circle $C_2$ has radius $r$ (where $1 < r < 9$) - Circle $C_3$ has radius $9$ - Point $A$ lies on $C_3$, so $OA = 9$ - Tangents from $A$ touch $C_2$ at points $B$ and $C$, so $OB = OC = r$ Step 2: Apply the tangency conditions. Since $AB$ and $AC$ are tangent lines from point $A$ to circle $C_2$: - $OB \perp AB$ (radius perpendicular to tangent) - $OC \perp AC$ (radius perpendicular to tangent) The line segment $BC$ is called the chord of contact, which is the polar line of point $A$ with respect to circle $C_2$. Step 3: Find the distance from $O$ to the chord $BC$. For a point $A$ outside a circle with center $O$ and radius $r$, the distance from $O$ to the chord of contact (polar line) is given by: $$d = \frac{r^2}{OA}$$ Substituting our values: $$d = \frac{r^2}{9}$$ Step 4: Use the condition that $BC$ is tangent to $C_1$. Since $BC$ is tangent to circle $C_1$ (which has radius $1$), the distance from center $O$ to the line $BC$ must equal the radius of $C_1$: $$\frac{r^2}{9} = 1$$ Step 5: Solve for $r$. From the equation above: $$r^2 = 9$$ $$r = 3$$ (We take the positive value since $r$ is a radius) Step 6: Verify the constraint and find the final answer. Check that $1 < r < 9$: Indeed, $1 < 3 < 9$ ✓ Therefore: $$10r = 10 \times 3 = 30$$ The answer is **Option 1: 30**.