To find the radius of the circle, let's analyze the problem step by step. We are given:

1. The circle passes through point $A(9, 0)$.
2. The circle is tangent to the $y$-axis at $B(0, 5)$.

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Step 1: Center of the Circle

The circle is tangent to the $y$-axis at $B(0, 5)$. Since the circle is tangent to the $y$-axis, the center of the circle must have coordinates $(r, 5)$, where $r$ is the radius (as the distance from the center to the tangent point is $r$).

Let the center of the circle be $C(r, 5)$.

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Step 2: Equation of the Circle

The equation of a circle with center $(h, k)$ and radius $r$ is given by: $(x - h)^2 + (y - k)^2 = r^2.$ Substituting $h = r$ and $k = 5$, the equation becomes: $(x - r)^2 + (y - 5)^2 = r^2.$

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Step 3: Substituting Point $A(9, 0)$

Since $A(9, 0)$ lies on the circle, it satisfies the circle equation: $(9 - r)^2 + (0 - 5)^2 = r^2.$

Simplify each term: $(9 - r)^2 = 81 - 18r + r^2, \quad (0 - 5)^2 = 25.$ Substitute these into the equation: $81 - 18r + r^2 + 25 = r^2.$

Cancel $r^2$ from both sides: $81 - 18r + 25 = 0.$

Combine constants: $106 - 18r = 0.$

Solve for $r$: $18r = 106, \quad r = \frac{106}{18} = \frac{53}{9}.$

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Final Answer:

The radius of the circle is: $\boxed{\frac{53}{9}}$

Let me know if you'd like further clarifications or have more questions! 😊

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Follow-Up Questions:

1. What is the general formula for a circle tangent to an axis?
2. How do we find the center of a circle given a tangent condition?
3. How do you solve for unknowns using substitution in circle equations?
4. Can you derive the equation of a circle tangent to the $x$-axis instead?
5. How does this problem change if $A$ and $B$ were swapped?

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Tip:

Always visualize the geometry and relationships in circle problems—tangency conditions often simplify complex setups!