Circles

Learning goals

Square the distance condition to get center-radius form
Complete the square once, with letters, to convert general form
Read the center as $\left(-\tfrac{D}{2}, -\tfrac{E}{2}\right)$
Let $D^2 + E^2 - 4F$ decide circle, point or nothing
Count line intersections with the discriminant
Classify two circles by the distance between centers

The distance condition

Fix a point $C(h, k)$ and a positive length $r$ . The circle with center $C$ and radius $r$ is the set of all points $P(x, y)$ whose distance from $C$ equals $r$ . Written as a sentence about distance, that is the whole definition:

PC = r.

To turn it into algebra you need the distance between two points in coordinates, and the grid hands you a right triangle for free. Travel from $C(h, k)$ straight across to the corner $(x, k)$ , then straight up to $P(x, y)$ . The across step has length $|x - h|$ , the up step has length $|y - k|$ , and they meet at a right angle, so the Pythagorean theorem gives the square of the direct path:

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

The absolute-value bars vanish under the squares, because squaring destroys a sign. That single line is the distance formula, and it is all the geometry this lesson needs.

A circle of radius 5 centered at C, at 1 and 2, with the point P, at 4 and 6, on it. Going across 3 units and then up 4 units builds a right triangle whose hypotenuse is the radius, so 3 squared plus 4 squared equals 5 squared. Every point of the circle sits at the end of such a triangle.

A point lies on the circle exactly when $(x - h)^2 + (y - k)^2 = r^2$ #

Let the circle have center $(h, k)$ and radius $r > 0$ , and let $P(x, y)$ be any point of the plane. By the definition, $P$ lies on the circle precisely when its distance to the center equals the radius, that is, when $PC = r$ .

Both $PC$ and $r$ are nonnegative numbers, and on nonnegative numbers squaring is reversible: if two nonnegative quantities are equal then their squares are equal. Conversely, if the squares of two nonnegative quantities are equal then the quantities themselves are equal, since each is the positive square root of its own square. So the statement $PC = r$ is equivalent to the statement $PC^2 = r^2$ , with nothing gained and nothing lost.

Now replace $PC^2$ by the Pythagorean expression computed above:

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

Every step was an equivalence, so the chain runs in both directions. A point satisfies this equation if and only if it sits exactly $r$ away from $(h, k)$ , which is exactly what it means to lie on the circle. The equation therefore captures every point of the curve and no other point of the plane.

This is the center-radius form (also called standard form) of a circle:

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

Read it carefully in both directions, because the form hides two traps. The form subtracts the center coordinates, so $(x + 3)$ means $h = -3$ , not $+3$ ; rewrite $x + 3$ as $x - (-3)$ whenever the sign is in doubt. And the number on the right is $r^2$ , not $r$ , so the radius of $(x - 1)^2 + (y - 2)^2 = 20$ is $\sqrt{20} = 2\sqrt{5}$ , not $20$ . Centered at the origin the form collapses to the tidy $x^2 + y^2 = r^2$ .

Worked example 1 A circle from the endpoints of a diameter

The points $(-2, 3)$ and $(4, 11)$ are the ends of a diameter. Find the circle.

The center is the midpoint of the diameter. Halfway along a segment you have covered half of its horizontal run and half of its vertical rise, so each midpoint coordinate is the average of the two endpoint coordinates:

(h, k) = \left( \frac{-2 + 4}{2}, \; \frac{3 + 11}{2} \right) = (1, 7).

The radius is half the diameter, so first square the distance between the endpoints. The horizontal gap is $4 - (-2) = 6$ and the vertical gap is $11 - 3 = 8$ :

d = \sqrt{6^2 + 8^2} = \sqrt{100} = 10, \qquad r = \tfrac{1}{2}(10) = 5.

Substituting $h = 1$ , $k = 7$ , and $r = 5$ into the center-radius form gives

(x - 1)^2 + (y - 7)^2 = 25.

Check it against an endpoint: $(-2 - 1)^2 + (3 - 7)^2 = 9 + 16 = 25$ , so $(-2, 3)$ is on the circle, as it must be. The classic error here is to use the whole distance $10$ as the radius, which builds a circle four times too big.

Completing the square once, with letters

Multiply out the center-radius form and the circle goes into hiding. Expanding $(x - h)^2 + (y - k)^2 = r^2$ gives $x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2$ , and moving everything to one side collects into

x^2 + y^2 + Dx + Ey + F = 0, \qquad D = -2h, \quad E = -2k, \quad F = h^2 + k^2 - r^2 .

This is the general form. The center and the radius are still in there, but they are scrambled across three coefficients. Recovering them means completing the square on the $x$ -terms and on the $y$ -terms separately, and that is a move you already own from the quadratics chapter. What is new is that you should do it once, with letters, and never again with numbers.

Completing the square converts $x^2 + y^2 + Dx + Ey + F = 0$ into center-radius form#

Group the $x$ -terms and the $y$ -terms, and move the constant to the right:

(x^2 + Dx) + (y^2 + Ey) = -F .

Complete each square. Half of the $x$ -coefficient is $\tfrac{D}{2}$ , and $\left(x + \tfrac{D}{2}\right)^2 = x^2 + Dx + \tfrac{D^2}{4}$ , so the group $x^2 + Dx$ is short of a perfect square by exactly $\tfrac{D^2}{4}$ . The same argument on the $y$ -group leaves it short by $\tfrac{E^2}{4}$ . Add both of those constants to both sides, which keeps the equation balanced:

\left(x^2 + Dx + \frac{D^2}{4}\right) + \left(y^2 + Ey + \frac{E^2}{4}\right) = -F + \frac{D^2}{4} + \frac{E^2}{4}.

Each group is now a perfect square, and the right side goes over the common denominator $4$ :

\left(x + \frac{D}{2}\right)^2 + \left(y + \frac{E}{2}\right)^2 = \frac{D^2 + E^2 - 4F}{4}.

Every general form lands here, whatever its coefficients. Comparing against the center-radius form, the center is $\left(-\tfrac{D}{2}, -\tfrac{E}{2}\right)$ , which is why the center coordinates are just half the linear coefficients with their signs flipped. The right side is whatever $\tfrac{D^2 + E^2 - 4F}{4}$ happens to be, and that number is $r^2$ only when it is positive.

The quantity $D^2 + E^2 - 4F$ does for a circle what $b^2 - 4ac$ does for a quadratic: its sign, and nothing else, decides what you are looking at. When it is positive, the equation is a genuine circle with

\text{center} \; \left(-\frac{D}{2},\, -\frac{E}{2}\right), \qquad r = \frac{\sqrt{D^2 + E^2 - 4F}}{2}.

Two conditions have to hold before you even reach for this. The coefficients of $x^2$ and $y^2$ must be equal and nonzero, and there must be no $xy$ term. If both squared terms carry the same coefficient $a \neq 1$ , as in $3x^2 + 3y^2 - 12x + 6y - 9 = 0$ , divide the whole equation by $a$ first and then proceed. If the coefficients differ, the graph is not a circle at all, and the lessons ahead in this chapter take up those curves.

Worked example 2 Convert $2x^2 + 2y^2 - 12x + 4y - 6 = 0$ to center-radius form

The squared terms have a common coefficient of $2$ , not $1$ , so divide every term by $2$ before doing anything else:

x^2 + y^2 - 6x + 2y - 3 = 0 .

Group and move the constant across, then complete both squares. Half of $-6$ is $-3$ (square $9$ ), and half of $2$ is $1$ (square $1$ ), so add $9$ and $1$ to both sides:

(x^2 - 6x + 9) + (y^2 + 2y + 1) = 3 + 9 + 1 .

Fold the groups back into squares and total the right side:

(x - 3)^2 + (y + 1)^2 = 13 .

The center is $(3, -1)$ and the radius is $\sqrt{13}$ , which is irrational and stays in radical form. The shortcut agrees: with $D = -6$ , $E = 2$ , $F = -3$ , the test value is $36 + 4 + 12 = 52$ , so the right side is $\tfrac{52}{4} = 13$ . Dividing by $2$ at the very start is the step everyone forgets, and skipping it corrupts every number that follows.

Check your understanding

Convert $x^2 + y^2 + 8x - 2y - 8 = 0$ to center-radius form. What are its center and radius?

A center

(-4, 1)

, radius

5

B center

(4, -1)

, radius

5

C center

(-4, 1)

, radius

25

D center

(-8, 2)

, radius

5

Circle, single point, or nothing at all

Completing the square always lands you at $(x - h)^2 + (y - k)^2 = C$ for some number $C$ on the right, and nothing forces $C$ to be positive. Squares of real numbers are never negative, so a sum of two of them behaves in three sharply different ways.

If $C > 0$ , the equation is an honest circle of radius $\sqrt{C}$ . If $C = 0$ , then a sum of two squares is zero, which happens only when each square is zero separately. That forces $x = h$ and $y = k$ , so the graph has collapsed to the single point $(h, k)$ . If $C < 0$ , the equation asks a sum of two squares to be negative, which no real point can do. So when $C < 0$ the graph is empty: there is no curve, no point, nothing to draw. The equation still has complex solutions, but nothing lands in the real plane you are graphing on.

Watch the three cases appear by turning one dial. Take $x^2 + y^2 - 4x + 6y + F = 0$ and vary only $F$ . The test value is $16 + 36 - 4F = 52 - 4F$ , so completing the square gives $(x - 2)^2 + (y + 3)^2 = 13 - F$ every time, and the center stays at $(2, -3)$ no matter what $F$ is. Raise $F$ and the circle shrinks: at $F = 9$ the radius is $2$ , and at $F = 12$ it is $1$ . At $F = 13$ the circle has closed down onto its own center, and past that it is gone.

One family, three outcomes. For the equation x squared plus y squared minus 4x plus 6y plus F equals 0, the center is always 2, negative 3. At F equals 9 the right side is 4 and the graph is a circle of radius 2. At F equals 13 the right side is 0 and the graph shrinks to the single point 2, negative 3. At F equals 20 the right side is negative 7 and no real point satisfies the equation.

So the claim “every equation of the form $x^2 + y^2 + Dx + Ey + F = 0$ is a circle” is simply false. A student who completes the square, reaches $(x - 2)^2 + (y + 3)^2 = -7$ , and announces a radius of $\sqrt{-7}$ has learned nothing about the graph. Check the sign of the right side before you claim a radius.

Worked example 3 For which $k$ is $x^2 + y^2 - 6x + 4y + k = 0$ a circle?

Here $D = -6$ , $E = 4$ , and $F = k$ , so the test value is

D^2 + E^2 - 4F = 36 + 16 - 4k = 52 - 4k .

Completing the square confirms it directly: $(x - 3)^2 + (y + 2)^2 = 13 - k$ , and indeed $13 - k = \tfrac{52 - 4k}{4}$ . The center is $(3, -2)$ for every $k$ , and only the right side moves.

The graph is a circle exactly when that right side is positive:

13 - k > 0 \iff k < 13 .

At $k = 13$ the right side is $0$ and the graph is the single point $(3, -2)$ . For $k > 13$ the right side is negative and there is no graph at all. The answer to the question asked is $k < 13$ , and the boundary case $k = 13$ is deliberately excluded, because a single point is not a circle.

Check your understanding

What is the graph of $x^2 + y^2 - 10x + 2y + 26 = 0$ ?

A a circle with center

(5, -1)

and radius

1

B the single point

(5, -1)

C a circle with center

(5, -1)

and radius

\sqrt{26}

D no points at all

Where a line meets a circle

A line and a circle can miss each other, touch at one point, or cut through at two. Which of the three happens is not a matter of drawing carefully: it is decided by a discriminant you already know how to compute.

Substitute the line $y = mx + c$ into the origin-centered circle $x^2 + y^2 = r^2$ and collect powers of $x$ :

x^2 + (mx + c)^2 = r^2 \;\Longrightarrow\; (1 + m^2)x^2 + 2mc\,x + (c^2 - r^2) = 0 .

A line meets a circle in at most two points#

The leading coefficient of that equation is $1 + m^2$ , and since $m^2 \geq 0$ for every real slope, we have $1 + m^2 \geq 1$ , which is never zero. So the substitution always produces a genuine quadratic in $x$ , never a linear equation in disguise, and a quadratic has at most two real roots.

Each root is one intersection point and no more, because the line assigns exactly one $y$ to each $x$ through $y = mx + c$ . Distinct roots therefore give distinct points, and there can be at most two of them.

Nothing in that argument used the center, so the same reasoning covers a circle sitting anywhere. Substituting the same line into $(x - h)^2 + (y - k)^2 = r^2$ and collecting powers of $x$ gives

(1 + m^2)x^2 + \left[\, 2m(c - k) - 2h \,\right] x + \left[\, h^2 + (c - k)^2 - r^2 \,\right] = 0,

whose leading coefficient is again $1 + m^2$ , never zero. A vertical line $x = a$ has no slope to substitute, but it behaves the same way. Substituting $x = a$ turns the circle into $(y - k)^2 = r^2 - (a - h)^2$ , which has two, one, or no real solutions for $y$ . Either way, no line can cross any circle three times.

Which of the three cases you get is settled by the discriminant of that quadratic, which for this origin-centered circle collapses to something tidy:

\Delta = (2mc)^2 - 4(1 + m^2)(c^2 - r^2) = 4\left[\, r^2(1 + m^2) - c^2 \,\right].

If $\Delta > 0$ there are two roots and the line is a secant, cutting the circle twice. If $\Delta = 0$ there is a repeated root and the line is a tangent, touching at exactly one point: tangency is nothing more exotic than a double root. If $\Delta < 0$ there are no real roots, and the line misses the circle entirely.

What generalizes is the method, not that closed form. The tidy expression above was computed from $x^2 + y^2 = r^2$ and holds only there. For a circle centered anywhere else, do not reach for it: substitute the line, collect the quadratic, and take the discriminant of the quadratic you actually have.

Three parallel lines against the circle x squared plus y squared equals 25. Substituting y equals 3 gives x squared equals 16, so x is 4 or negative 4 and the line is a secant. Substituting y equals 5 gives x squared equals 0, a double root, so the line is tangent at 0, 5. Substituting y equals 7 gives x squared equals negative 24, which has no real solution, so the line misses the circle.

Worked example 4 Where does $y = x + 1$ meet $x^2 + y^2 = 25$ ?

Substitute the line into the circle, replacing $y$ everywhere it appears:

x^2 + (x + 1)^2 = 25 \;\Longrightarrow\; 2x^2 + 2x - 24 = 0 \;\Longrightarrow\; x^2 + x - 12 = 0 .

This factors as $(x + 4)(x - 3) = 0$ , so $x = -4$ or $x = 3$ . Two real roots means the line is a secant.

Now finish the job. A value of $x$ is not a point; feed each root back into the line to get its partner:

x = -4 \Rightarrow y = -3, \qquad x = 3 \Rightarrow y = 4 .

The intersection points are $(-4, -3)$ and $(3, 4)$ . Both check out on the circle, since $16 + 9 = 25$ and $9 + 16 = 25$ . Stopping at the $x$ -values is the most common way to lose this problem.

Worked example 5 Which lines $y = x + c$ are tangent to $x^2 + y^2 = 25$ ?

Tangency means the substituted quadratic has a repeated root, so build the quadratic and set its discriminant to zero. With slope $m = 1$ :

x^2 + (x + c)^2 = 25 \;\Longrightarrow\; 2x^2 + 2cx + (c^2 - 25) = 0 .

Its discriminant is $(2c)^2 - 4(2)(c^2 - 25)$ , which simplifies to $-4c^2 + 200$ . Setting that to zero:

-4c^2 + 200 = 0 \;\Longrightarrow\; c^2 = 50 \;\Longrightarrow\; c = \pm 5\sqrt{2}.

Two lines qualify, $y = x + 5\sqrt{2}$ and $y = x - 5\sqrt{2}$ , one riding above the circle and one below. That pairing is exactly what the picture demands: a circle has two tangent lines of any given slope. The origin-centered formula $\Delta = 4[r^2(1 + m^2) - c^2]$ agrees, since $r^2(1 + m^2) = 25(2) = 50$ , and it applies here only because this circle is centered at the origin. Move the center and that shortcut stops being true, so substitute and recompute instead.

Check your understanding

How many points does the line $y = x + 8$ share with the circle $x^2 + y^2 = 25$ ?

A two B one C none D infinitely many

Where two circles meet

Two circles are settled the same way, by a single number: the distance $d$ between their centers, weighed against the radii $r_1$ and $r_2$ . Picture sliding one ring toward the other and watching the contact change.

While $d > r_1 + r_2$ the rings cannot reach each other, so they share no points. At $d = r_1 + r_2$ they reach exactly and touch at one point, tangent from the outside. Closer still, whenever $|r_1 - r_2| < d < r_1 + r_2$ , they overlap and cross at two points. At $d = |r_1 - r_2|$ , with radii that differ, the smaller ring rests inside the larger and touches it once from within. And once $d < |r_1 - r_2|$ the smaller circle is swallowed whole, floating strictly inside the larger with no contact at all.

So two circles can fail to meet for two opposite reasons: they can be too far apart, or one can be buried inside the other. Closer does not always mean more contact, which is the detail most people miss.

When two circles do cross, their intersection points are cheap to find, because both equations carry the same $x^2 + y^2$ . Subtract one general form from the other and those squared terms cancel, leaving a linear equation, a line. Any point on both circles satisfies both equations, so it satisfies their difference, which means every shared point lies on that line. Solve the line against either circle and you have them. It also settles the count: a line meets a circle at most twice, so two distinct circles share at most two points.

Three points pin down one circle

The general form looks uglier than the center-radius form, but it earns its keep in one situation the tidy form cannot handle. Look again at

x^2 + y^2 + Dx + Ey + F = 0 .

If you treat $D$ , $E$ , and $F$ as the unknowns and a known point $(x_0, y_0)$ as the data, this equation is linear in those unknowns. This is because $D$ , $E$ , and $F$ each appear to the first power, multiplied by constants. Each point the circle must pass through gives one linear equation. Three points give three linear equations in three unknowns, which is a system you already know how to solve.

Worked example 6 The circle through $(1, 1)$ , $(2, 4)$ , and $(5, 3)$

Substitute each point into $x^2 + y^2 + Dx + Ey + F = 0$ and simplify, keeping $D$ , $E$ , and $F$ as the unknowns:

\begin{aligned} (1, 1) &: \quad D + E + F = -2, \\ (2, 4) &: \quad 2D + 4E + F = -20, \\ (5, 3) &: \quad 5D + 3E + F = -34 . \end{aligned}

Eliminate $F$ by subtracting the first equation from the other two, which leaves a $2 \times 2$ system:

D + 3E = -18, \qquad 4D + 2E = -32 .

The second simplifies to $2D + E = -16$ . Solving the first for $D = -18 - 3E$ and substituting gives $2(-18 - 3E) + E = -16$ , so $-5E = 20$ and $E = -4$ . Back-substituting, $D = -18 + 12 = -6$ , and then $F = -2 - D - E = -2 + 6 + 4 = 8$ . The circle is

x^2 + y^2 - 6x - 4y + 8 = 0 .

Complete the square to see it: $(x - 3)^2 + (y - 2)^2 = -8 + 9 + 4 = 5$ , a circle with center $(3, 2)$ and radius $\sqrt{5}$ . Verify with the data you were given. The squared distance from $(3, 2)$ to $(1, 1)$ is $4 + 1 = 5$ , to $(2, 4)$ is $1 + 4 = 5$ , and to $(5, 3)$ is $4 + 1 = 5$ . All three points sit exactly $\sqrt{5}$ from the center, so all three are on the circle.

The same linearity explains the geometry: three unknowns need three conditions, so three points determine the circle, no more and no fewer. Two points leave a whole family of circles to choose from, and four points are usually one demand too many for any circle to satisfy. The one case that fails is three points on a straight line, where the elimination collapses and no circle exists.

Common mistakes

Flipping the sign of the center. The form subtracts, so $(x + 5)^2 + (y - 2)^2 = 9$ has center $(-5, 2)$ . Rewrite $x + 5$ as $x - (-5)$ whenever you are unsure.
Reporting the right side as the radius. The right side is $r^2$ . A right side of $20$ means $r = \sqrt{20} = 2\sqrt{5}$ , not $20$ .
Forgetting to divide by the leading coefficient. In $3x^2 + 3y^2 - 12x + 6y - 9 = 0$ you must divide the whole equation by $3$ before completing the square. The center and radius formulas assume the squared terms have coefficient $1$ .
Completing the square on one side only. The constants you add to complete each square must be added to the right side too, or the equation stops being true.
Announcing a radius without checking the sign. If completing the square leaves $(x - 2)^2 + (y + 3)^2 = -7$ , there is no radius and no graph. A negative right side means the real plane holds no solutions at all, and a right side of $0$ means a single point, not a circle.
Assuming every second-degree equation is a circle. A circle needs equal, nonzero coefficients on $x^2$ and $y^2$ , and no $xy$ term. Unequal coefficients describe some other curve, or a degenerate case with no points at all (as in $x^2 + 2y^2 = -1$ ), but never a circle.
Confusing a diameter with a radius. Given the two endpoints of a diameter, the radius is half the distance between them. Given the center and one point on the circle, the radius is the whole distance.
Stopping at the $x$ -values. When a line meets a circle, solving the quadratic gives $x$ -values. Substitute each back into the line to produce an actual point $(x, y)$ .

Key takeaways

A circle is the set of points at a fixed distance $r$ from a fixed center $(h, k)$ . Squaring that distance condition gives the center-radius form $(x - h)^2 + (y - k)^2 = r^2$ , and the equation holds for a point if and only if that point is on the circle.
Expanding gives the general form $x^2 + y^2 + Dx + Ey + F = 0$ , and completing the square once, with letters, converts it back.
The result of that one computation is $\left(x + \tfrac{D}{2}\right)^2 + \left(y + \tfrac{E}{2}\right)^2 = \tfrac{D^2 + E^2 - 4F}{4}$ , so the center is always $\left(-\tfrac{D}{2}, -\tfrac{E}{2}\right)$ .
The sign of $D^2 + E^2 - 4F$ decides everything: positive gives a circle of radius $\tfrac{1}{2}\sqrt{D^2 + E^2 - 4F}$ , zero gives the single point $(h, k)$ , and negative gives no real points at all.
A circle’s equation has equal, nonzero coefficients on $x^2$ and $y^2$ and no $xy$ term. If that shared coefficient is not $1$ , divide the equation by it first.
Substituting a line into a circle always yields a genuine quadratic, so a line meets a circle in at most two points. The discriminant counts them: positive gives a secant, zero gives a tangent (a double root), and negative gives no intersection.
Two circles are counted by the distance $d$ between their centers. They share no points when $d > r_1 + r_2$ or when $d < |r_1 - r_2|$ (one circle buried inside the other), exactly one point when $d$ equals either bound, and two points strictly between them. Subtracting their general forms cancels the squared terms and leaves the line through both intersection points.
The general form is linear in $D$ , $E$ , and $F$ , so three points on the circle give three linear equations, and solving that system produces the unique circle through them.

Circles

Learning goals

The distance condition

Completing the square once, with letters

Circle, single point, or nothing at all

Where a line meets a circle

Where two circles meet

Three points pin down one circle

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)