How to write the equation of a circle from tangent lines

Key takeaway: This is one of those coordinate-geometry problems where the tangents do most of the work. If a circle touches two vertical lines and the x-axis, the center is forced into a very specific spot.

A circle tangent to two vertical lines has its center halfway between them.

Step 1: Find the x-coordinate of the center

The circle is tangent to $x=-12$ and $x=-4$, so the center’s x-coordinate is the midpoint:

$\frac{-12+(-4)}{2}=-8$

Step 2: Find the radius

The distance from the center to either vertical tangent line is the radius:

$r=|-8-(-12)|=4$

Step 3: Find the y-coordinate of the center

Because the circle is tangent to the x-axis, the center is 4 units above it in Quadrant II:

$y=4$

So the center is

$(-8,4)$

and the radius is

$4$

Step 4: Write the equation

Use standard form:

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

Substitute $h=-8$, $k=4$, and $r=4$:

$(x+8)^2+(y-4)^2=16$

That is the equation of the circle.

---

Pitfalls the pros know 👇 A common mistake is treating tangency like intersection. The circle does not cross those lines; it just touches them once, so the distance from the center to each tangent line equals the radius. Another slip is forgetting that the x-axis tangent means the center sits one radius above $y=0$.

What if the problem changes? If the circle were tangent to $x=a$ and $x=b$ instead, the center’s x-coordinate would still be the midpoint $\frac{a+b}{2}$. If it were tangent to the y-axis rather than the x-axis, you would use the same idea vertically and write the center one radius away from $x=0$ instead.

Tags: tangent lines, center of a circle, standard form