Introduction to Circle Geometry
In coordinate geometry, the equation of a circle is defined by its center $(h, k)$ and its radius $r$. The standard form of the equation is given by:
$(x - h)^2 + (y - k)^2 = r^2$
To solve the problem provided, we need to understand how the condition of a circle "touching the x-axis" relates to its radius.
The Concept: Tangency to the Axis
When a circle touches the x-axis, the perpendicular distance from the center $(h, k)$ to the x-axis is exactly equal to the radius $r$.
- The x-axis is defined by the line $y = 0$.
- The distance from any point $(h, k)$ to the line $y = 0$ is simply the absolute value of the y-coordinate, $|k|$.
- Therefore, for a circle touching the x-axis, $r = |k|$.
Solving the Problem
Given the center $(h, k) = (-3, -4)$, we can determine the radius:
Identify the radius: Since the circle touches the x-axis, $r = |k|$. Here, $k = -4$, so: $r = |-4| = 4$
State the standard form: We use the formula $(x - h)^2 + (y - k)^2 = r^2$. Substituting $h = -3$, $k = -4$, and $r = 4$: $(x - (-3))^2 + (y - (-4))^2 = 4^2$
Simplify the equation: $(x + 3)^2 + (y + 4)^2 = 16$
Expand into general form (optional): $(x^2 + 6x + 9) + (y^2 + 8y + 16) = 16$ $x^2 + y^2 + 6x + 8y + 9 = 0$
Summary
The equation of the circle with center $(-3, -4)$ that is tangent to the x-axis is $(x + 3)^2 + (y + 4)^2 = 16$. Always remember: when a circle touches the x-axis, its radius is the absolute value of its y-coordinate. Conversely, if it touches the y-axis, its radius is the absolute value of its x-coordinate.