Math • April 2026

Finding the Equation of
a Circle Touching the X-axis

A
Written By Archive Editorial
Reading Time 5 Min Read

Understanding the Problem

To find the equation of a circle, we typically need two pieces of information: the coordinates of its center $(h, k)$ and its radius $r$. The standard form of a circle's equation is:

$(x - h)^2 + (y - k)^2 = r^2$

In this problem, we are given:

  • The center $(h, k) = (-3, -4)$
  • The condition that the circle touches the x-axis.

Visualizing the Geometry

When a circle touches the x-axis, the perpendicular distance from the center $(h, k)$ to the x-axis must be equal to the radius of the circle.

  • The x-axis is the line $y = 0$.
  • The distance from any point $(h, k)$ to the x-axis is simply the absolute value of its y-coordinate, $|k|$.
  • Therefore, $r = |k|$.

Given the center $(-3, -4)$, the radius is: $r = |-4| = 4$

Step-by-Step Solution

1. Identify the center and radius

  • Center $(h, k) = (-3, -4)$
  • Radius $r = 4$

2. Substitute into the standard equation

Using the formula $(x - h)^2 + (y - k)^2 = r^2$, we plug in our values: $(x - (-3))^2 + (y - (-4))^2 = 4^2$

3. Simplify the equation

$(x + 3)^2 + (y + 4)^2 = 16$

4. Optional: Expand to General Form

If you need the general form, expand the squares: $(x^2 + 6x + 9) + (y^2 + 8y + 16) = 16$ $x^2 + y^2 + 6x + 8y + 25 = 16$ $x^2 + y^2 + 6x + 8y + 9 = 0$

Conclusion

The equation of the circle is $(x + 3)^2 + (y + 4)^2 = 16$ (or $x^2 + y^2 + 6x + 8y + 9 = 0$).

Pro-Tip: If a circle touches the y-axis instead, the radius would be the absolute value of the x-coordinate, $r = |h|$. Always check which axis the circle is tangent to!

Editorial Note

Automated explainer generated from student question.

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