Key Concepts and Formulas

• General Equation of a Circle: A circle with center $(h, k)$ and radius $r$ has the equation: $(x - h)^2 + (y - k)^2 = r^2$
• Circle Tangent to the x-axis: If a circle is tangent to the x-axis, the distance from its center $(h, k)$ to the x-axis is equal to its radius. Therefore, the radius $r = |k|$.
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step-by-Step Solution

Step 1: Define the equation of the circle. Since the circle is tangent to the x-axis, its radius is $|k|$. Thus, the equation of the circle with center $(h, k)$ is: $(x - h)^2 + (y - k)^2 = k^2$ We use $k^2$ instead of $|k|^2$ because they are equivalent.

Step 2: Use the given point to find a relationship between $h$ and $k$. The circle passes through the point $(-1, 1)$. Substitute $x = -1$ and $y = 1$ into the equation of the circle: $(-1 - h)^2 + (1 - k)^2 = k^2$

Step 3: Simplify the equation. Expand and simplify the equation to find a relationship between $h$ and $k$: $(1 + 2h + h^2) + (1 - 2k + k^2) = k^2$ $1 + 2h + h^2 + 1 - 2k + k^2 - k^2 = 0$ $h^2 + 2h + 2 - 2k = 0$ $2k = h^2 + 2h + 2$

Step 4: Express $k$ in terms of $h$. Solve for $k$ to express it as a function of $h$: $k = \frac{h^2 + 2h + 2}{2}$

Step 5: Complete the square. Complete the square in the expression for $k$ to find its minimum value: $k = \frac{h^2 + 2h + 1 + 1}{2}$ $k = \frac{(h + 1)^2 + 1}{2}$ $k = \frac{(h + 1)^2}{2} + \frac{1}{2}$

Step 6: Determine the range of $k$. Since $(h + 1)^2 \ge 0$ for all real numbers $h$, we have: $\frac{(h + 1)^2}{2} \ge 0$ Therefore, $k = \frac{(h + 1)^2}{2} + \frac{1}{2} \ge \frac{1}{2}$ So, $k \ge \frac{1}{2}$.

Step 7: Reconsider the problem statement and the tangency condition. The original solution incorrectly states that the correct answer is $- {1 \over 2} \le k \le {1 \over 2}$. However, from the previous steps, we derived that $k \ge \frac{1}{2}$. Since the question mentions that the circle passes through $(-1, 1)$ and is tangent to the x-axis, it must be the case that $k$ must be greater than 0. If $k$ were negative, the circle would lie below the x-axis, and passing through $(-1, 1)$ would not be possible while also being tangent to the x-axis.

The problem statement is incorrect. If we consider the condition that the circle passes through $(-1,1)$ and is tangent to the x-axis, we get the result $k \ge \frac{1}{2}$.

If the problem statement meant that $-1 \le h \le 1$, then the answer would be $- {1 \over 2} \le k \le {1 \over 2}$. If $h = -1$, $k = \frac{(-1)^2 + 2(-1) + 2}{2} = \frac{1 - 2 + 2}{2} = \frac{1}{2}$ If $h = 1$, $k = \frac{(1)^2 + 2(1) + 2}{2} = \frac{1 + 2 + 2}{2} = \frac{5}{2}$

If we assume there is a typo and the point is $(-1, 0)$, then: $(-1 - h)^2 + (0 - k)^2 = k^2$ $1 + 2h + h^2 + k^2 = k^2$ $1 + 2h + h^2 = 0$ $(h + 1)^2 = 0$ $h = -1$ Then $k$ can be any value.

Step 8: Assume the correct answer is indeed option (D) $k \ge {1 \over 2}$. We have already derived that $k \ge \frac{1}{2}$.

Common Mistakes & Tips

• Remember that the radius of the circle is the absolute value of the y-coordinate of the center when the circle is tangent to the x-axis.
• Completing the square is a useful technique for finding the minimum or maximum value of a quadratic expression.
• Always check your answer to see if it makes sense in the context of the problem.

Summary

We started by defining the equation of a circle tangent to the x-axis and passing through the point $(-1, 1)$. We then substituted the coordinates of the point into the equation and simplified to find a relationship between $h$ and $k$. After completing the square, we found that $k \ge \frac{1}{2}$. Therefore, the correct set of values for $k$ is $k \ge \frac{1}{2}$.

Final Answer

The final answer is \boxed{k \ge {1 \over 2}}, which corresponds to option (D).