Circles in the Plane

A circle is a set, and sets have equations

Definition. The circle with center $(h, k)$ and radius $r > 0$ is the set

$\{(x, y) \in \mathbb{R}^2 : \text{distance from } (x,y) \text{ to } (h,k) = r\}.$

We know how to measure that distance. The Pythagorean theorem (or its coordinate avatar, the distance formula) gives:

$\sqrt{(x-h)^2 + (y-k)^2} = r.$

Square both sides:

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

That's the standard form of a circle's equation. It was not handed down from the sky, it was not memorized, it was not in some table in the back of a textbook — we derived it from the definition of "circle" using the distance formula. The equation IS the definition, in algebra form. No bullshit.

Special case: center at the origin. Set $h = k = 0$:

$x^2 + y^2 = r^2.$

This is the unit circle's equation when $r = 1$: $x^2 + y^2 = 1$. You've already been using it in the trig identities lesson.

Reading a circle from standard form

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

• Center $= (h, k)$.
• Radius $= \sqrt{r^2} = r$ (positive).

The most common trap: the signs. $(x - 3)^2 + (y + 2)^2 = 25$ has center $(3, -2)$ — the $+2$ in $(y+2)$ means $k = -2$. Write it as $(y - (-2))^2$ and the pattern is clear.

Completing the square to find the circle

When a circle equation is given in general (expanded) form

$x^2 + y^2 + Dx + Ey + F = 0,$

we use completing the square to get it into standard form. This is the same algebraic procedure from the Quadratic Equations lesson — now in two variables simultaneously.

Procedure:

1. Group $x$-terms and $y$-terms; move $F$ to the right.
2. Complete the square in $x$: add $(D/2)^2$ to both sides.
3. Complete the square in $y$: add $(E/2)^2$ to both sides.
4. Write as $(x-h)^2 + (y-k)^2 = r^2$ and read off center and radius.

Example. $x^2 + y^2 - 6x + 4y - 12 = 0$.

Group: $(x^2 - 6x) + (y^2 + 4y) = 12$.

Complete in $x$: $(-6/2)^2 = 9$. Add to both sides: $(x^2 - 6x + 9) + (y^2 + 4y) = 12 + 9 = 21$.

Complete in $y$: $(4/2)^2 = 4$. Add to both sides: $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 21 + 4 = 25$.

Factor: $(x - 3)^2 + (y + 2)^2 = 25$.

Center $= (3, -2)$, radius $= 5$.

Notice: completing the square is doing its third job in this course. First job: solving quadratics. Second job: converting parabolas to vertex form. Third job: reading circles. Same algebraic move, three different contexts — one tool, many applications, zero apologies. I love this tool. It's the Swiss Army knife of algebra and the Federation of Boring Textbook Authors treats it like a footnote.

Tangent lines: perpendicular to radii

A tangent line to a circle touches the circle at exactly one point. The key geometric fact:

Theorem. A tangent to a circle at a point $P$ is perpendicular to the radius at $P$.

Proof. The tangent line intersects the circle at one point only. For any OTHER point $Q$ on the tangent line, $Q$ is outside the circle, so its distance to the center $O$ is greater than $r$. The shortest path from $O$ to the line is the perpendicular — and the perpendicular from $O$ to the tangent hits it at $P$ (since $OP = r$ is the minimum distance). Therefore the radius $OP$ is perpendicular to the tangent. $\square$

Finding a tangent line

Given a circle with center $O = (h, k)$ and a point $P = (x_0, y_0)$ on the circle, the tangent at $P$ is perpendicular to the radius $OP$.

1. Slope of radius $OP$: $m_{OP} = \frac{y_0 - k}{x_0 - h}$.
2. Slope of tangent: $m_T = -\frac{1}{m_{OP}}$ (negative reciprocal, because perpendicular).
3. Tangent line: $y - y_0 = m_T(x - x_0)$.

Intersecting a line and a circle

To find where a line meets a circle, substitute the line equation into the circle equation and solve the resulting quadratic. The discriminant of that quadratic tells you the story:

• Two solutions ($\Delta > 0$): the line is a secant (cuts through the circle at two points).
• One solution ($\Delta = 0$): the line is tangent (touches at exactly one point).
• No real solutions ($\Delta < 0$): the line misses the circle entirely.

Example. Does the line $y = x + 3$ intersect the circle $x^2 + y^2 = 25$?

Substitute: $x^2 + (x+3)^2 = 25$, so $x^2 + x^2 + 6x + 9 = 25$, giving $2x^2 + 6x - 16 = 0$, or $x^2 + 3x - 8 = 0$.

Discriminant: $\Delta = 9 + 32 = 41 > 0$. Two intersections — the line is a secant.

The quadratic machinery from Algebra Core is doing its job here: the intersection question reduces to "does this equation have real roots?" and the discriminant answers it in one shot.