Key Concepts and Formulas

• Standard Equation of a Circle: $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.
• Circle Touching the x-axis: If a circle touches the x-axis at $(a, 0)$, its center is $(a, r)$ or $(a, -r)$ and the radius is $|r|$.
• Intercept on the y-axis: If a circle intercepts the y-axis, the length of the intercept is $2\sqrt{r^2 - h^2}$, where $h$ is the x-coordinate of the center.

Step-by-Step Solution

Step 1: Define the center and radius based on the x-axis tangency.

Since the circle touches the x-axis at $(3, 0)$, the x-coordinate of the center is 3. Let the center be $(3, k)$ and the radius be $|k|$. Thus, the equation of the circle is $(x-3)^2 + (y-k)^2 = k^2$.

Step 2: Use the y-axis intercept information.

The circle makes an intercept of length 8 on the y-axis. This means the circle intersects the y-axis at two points, say $(0, y_1)$ and $(0, y_2)$, such that $|y_1 - y_2| = 8$. Substituting $x = 0$ into the circle's equation, we get: $(0-3)^2 + (y-k)^2 = k^2$ $9 + y^2 - 2ky + k^2 = k^2$ $y^2 - 2ky + 9 = 0$ This is a quadratic equation in $y$. The roots of this equation are $y_1$ and $y_2$.

Step 3: Relate the roots of the quadratic to the intercept length.

The difference between the roots is given by $|y_1 - y_2| = \sqrt{(y_1 + y_2)^2 - 4y_1y_2}$. From the quadratic equation, the sum of the roots $y_1 + y_2 = 2k$ and the product of the roots $y_1y_2 = 9$. Thus, $|y_1 - y_2| = \sqrt{(2k)^2 - 4(9)} = 8$ $\sqrt{4k^2 - 36} = 8$ $4k^2 - 36 = 64$ $4k^2 = 100$ $k^2 = 25$ $k = \pm 5$

Step 4: Determine the two possible equations of the circle.

We have two possible centers: $(3, 5)$ and $(3, -5)$.

• If the center is $(3, 5)$, the radius is 5, and the equation of the circle is $(x-3)^2 + (y-5)^2 = 25$.
• If the center is $(3, -5)$, the radius is 5, and the equation of the circle is $(x-3)^2 + (y+5)^2 = 25$.

Step 5: Test the given points to see which lies on either of the circles.

• (A) (1, 5):

  • For $(x-3)^2 + (y-5)^2 = 25$: $(1-3)^2 + (5-5)^2 = (-2)^2 + 0^2 = 4 \neq 25$.
  • For $(x-3)^2 + (y+5)^2 = 25$: $(1-3)^2 + (5+5)^2 = (-2)^2 + (10)^2 = 4 + 100 = 104 \neq 25$. There must be an error. Let's re-examine the y-intercept. The y intercepts are $y = k \pm \sqrt{k^2 - 9}$. The length of the intercept is $2\sqrt{k^2-9}=8$, so $\sqrt{k^2-9}=4$, $k^2 - 9 = 16$, $k^2 = 25$, $k=\pm 5$. The equations are $(x-3)^2 + (y-5)^2 = 25$ and $(x-3)^2 + (y+5)^2 = 25$.

  Now let's test the points again.

  • (A) (1, 5):
    • $(1-3)^2 + (5-5)^2 = 4 + 0 = 4 \ne 25$
    • $(1-3)^2 + (5+5)^2 = 4 + 100 = 104 \ne 25$
  • (B) (2, 3):
    • $(2-3)^2 + (3-5)^2 = 1+4 = 5 \ne 25$
    • $(2-3)^2 + (3+5)^2 = 1 + 64 = 65 \ne 25$
  • (C) (3, 5):
    • $(3-3)^2 + (5-5)^2 = 0 \ne 25$
    • $(3-3)^2 + (5+5)^2 = 100 \ne 25$
  • (D) (3, 10):
    • $(3-3)^2 + (10-5)^2 = 25$. So, (3, 10) lies on the circle $(x-3)^2 + (y-5)^2 = 25$.
    • $(3-3)^2 + (10+5)^2 = 225 \ne 25$.

There appears to be an error in the given answer. Let's re-evaluate.

The equations are: $(x-3)^2 + (y-5)^2 = 25$ and $(x-3)^2 + (y+5)^2 = 25$.

Checking option (A) (1,5): $(1-3)^2 + (5-5)^2 = 4 + 0 = 4 \neq 25$ $(1-3)^2 + (5+5)^2 = 4 + 100 = 104 \neq 25$

Let's re-examine the y-intercept calculation. We have $y^2 - 2ky + 9 = 0$. The y-intercept length is 8. The y-intercepts are $k \pm \sqrt{k^2-9}$. Then $k + \sqrt{k^2-9} - (k - \sqrt{k^2-9}) = 2\sqrt{k^2-9} = 8$. So $\sqrt{k^2-9} = 4$ and $k^2-9=16$ and $k^2 = 25$, so $k = \pm 5$.

Let's carefully re-check the options.

(A) (1, 5): $(1-3)^2 + (5-5)^2 = 4 \ne 25$ and $(1-3)^2 + (5+5)^2 = 4+100 = 104 \ne 25$ (B) (2, 3): $(2-3)^2 + (3-5)^2 = 1+4 = 5 \ne 25$ and $(2-3)^2 + (3+5)^2 = 1+64 = 65 \ne 25$ (C) (3, 5): $(3-3)^2 + (5-5)^2 = 0 \ne 25$ and $(3-3)^2 + (5+5)^2 = 100 \ne 25$ (D) (3, 10): $(3-3)^2 + (10-5)^2 = 25$ and $(3-3)^2 + (10+5)^2 = 225 \ne 25$

So, (3, 10) lies on the circle $(x-3)^2 + (y-5)^2 = 25$. The correct option should be (D), but it is given as (A). However, the question asks for a point that passes through the circle.

If $k=5$, $(x-3)^2 + (y-5)^2 = 25$. If $k=-5$, $(x-3)^2 + (y+5)^2 = 25$.

Let's reconsider the options.

(A) (1,5): $(1-3)^2 + (5-5)^2 = 4 \neq 25$. $(1-3)^2 + (5+5)^2 = 104 \neq 25$. (B) (2,3): $(2-3)^2 + (3-5)^2 = 5 \neq 25$. $(2-3)^2 + (3+5)^2 = 65 \neq 25$. (C) (3,5): $(3-3)^2 + (5-5)^2 = 0 \neq 25$. $(3-3)^2 + (5+5)^2 = 100 \neq 25$. (D) (3,10): $(3-3)^2 + (10-5)^2 = 25$. $(3-3)^2 + (10+5)^2 = 225 \neq 25$.

Only (3, 10) fits for $k=5$, which corresponds to $(x-3)^2 + (y-5)^2 = 25$.

Common Mistakes & Tips

• Always double-check your algebraic manipulations, especially when dealing with square roots and quadratic equations.
• Make sure you understand the geometric implications of each condition given in the problem.
• When dealing with intercepts, remember the relationship between the roots of a quadratic equation and its coefficients.

Summary

We determined the equation of the circle using the tangency condition and the y-intercept length. We found two possible equations for the circle. By substituting the coordinates of the given points into these equations, we determined which point lies on one of the circles. Point (3, 10) satisfies the equation $(x-3)^2 + (y-5)^2 = 25$. However, the given answer is (1,5) which is incorrect. Upon review, there appears to be a mistake in the options, and the point (3,10) is the correct solution. There is clearly an error in the provided correct answer.

Final Answer

The final answer is \boxed{(3, 10)}. Option (D) is the correct answer.