Equation of Circles Touching X Axis with Given Y Intercept

To find the equation of the circle(s), we use the geometric properties given in the problem:

1. Condition 1: Touching the $x$-axis at a distance of $3$ from the origin Based on the provided options, the circle touches the positive $x$-axis at the point $(3, 0)$. Since the circle is tangent to the $x$-axis at $(3, 0)$, the center of the circle must have an $x$-coordinate of $h = 3$, and its radius $R$ is equal to the absolute value of the $y$-coordinate of the center, i.e., $R = |k|$.

   Therefore, the center of the circle is $(3, k)$ and its radius is $R = |k|$. The standard equation of the circle is: $(x - 3)^2 + (y - k)^2 = k^2$ Expanding this equation gives: $x^2 - 6x + 9 + y^2 - 2ky + k^2 = k^2$ $x^2 + y^2 - 6x - 2ky + 9 = 0 \quad \text{--- (1)}$

2. Condition 2: Intercept on the $y$-axis has length $2\sqrt{7}$ To find the $y$-intercepts, we substitute $x = 0$ into equation (1): $y^2 - 2ky + 9 = 0$ Let the roots of this quadratic equation be $y_1$ and $y_2$. By Vieta's formulas: $y_1 + y_2 = 2k$ $y_1 y_2 = 9$

   The length of the intercept made on the $y$-axis is: $|y_1 - y_2| = \sqrt{(y_1 + y_2)^2 - 4y_1 y_2} = \sqrt{(2k)^2 - 4(9)} = 2\sqrt{k^2 - 9}$

   Given that the length of the intercept is $2\sqrt{7}$: $2\sqrt{k^2 - 9} = 2\sqrt{7}$ $\sqrt{k^2 - 9} = \sqrt{7}$ Squaring both sides: $k^2 - 9 = 7 \implies k^2 = 16 \implies k = \pm 4$

3. Forming the Circle Equations:

   • For $k = 4$: $x^2 + y^2 - 6x - 2(4)y + 9 = 0 \implies x^2 + y^2 - 6x - 8y + 9 = 0$
   • For $k = -4$: $x^2 + y^2 - 6x - 2(-4)y + 9 = 0 \implies x^2 + y^2 - 6x + 8y + 9 = 0$

Thus, the possible equations of the circles are:

• $x^2 + y^2 - 6x + 8y + 9 = 0$ (Option A)
• $x^2 + y^2 - 6x - 8y + 9 = 0$ (Option C)

Correct Options: A, C