Circles

Learning goals

Square the distance condition to get center-radius form
Complete the square to convert a circle's general form into center-radius form
Read the center and classify circle, point, or nothing from $D$ , $E$ , and $F$
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.

Here is one such circle: center $C(1, 2)$ , radius $5$ , with a point $P(4, 6)$ on it.

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.

To turn “exactly $r$ away” into algebra, use the right triangle the grid hands you for free. Travel from $C(h, k)$ straight across to the corner $(x, k)$ , then straight up to $P(x, y)$ , the way the diagram above does it for $C(1, 2)$ and $P(4, 6)$ : across $3$ , up $4$ , and $3^2 + 4^2 = 5^2$ . 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 for any center and point, not just this one:

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 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 exactly when $PC = r$ .

$PC$ and $r$ are both nonnegative, and squaring a nonnegative number is reversible: two nonnegative quantities are equal exactly when their squares are equal, since each number is the positive square root of its own square. So $PC = r$ says exactly the same thing as $PC^2 = r^2$ .

Replace $PC^2$ with the Pythagorean expression from above:

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

Because every step was reversible, this equation holds for a point exactly when that point sits $r$ away from $(h, k)$ , that is, exactly when the point lies on the circle.

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

Check your understanding

A circle has center $(3, -2)$ and radius $7$ . Squaring the distance from a point $(x, y)$ to that center gives which center-radius equation?

(x - 3)^2 + (y + 2)^2 = 49

(x + 3)^2 + (y - 2)^2 = 49

(x - 3)^2 + (y + 2)^2 = 7

(x - 3) + (y + 2) = 49

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: that doubles the radius, which makes the area four times as large.

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. Try it once on a specific equation, then once more with letters, so the letter version hands you a ready-made shortcut for every equation after this one.

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. Dividing by $2$ at the very start is the step everyone forgets, and skipping it corrupts every number that follows.

The steps above work on any equation of that shape. Repeat them with letters instead of numbers, and the pattern turns into a shortcut you can read off without redoing the arithmetic by hand.

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.

Two conditions have to hold before you reach for that shortcut. 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, exactly as Worked Example 2 did. If the coefficients differ, the graph is not a circle at all, and the lessons ahead in this chapter take up those curves.

Once those conditions hold, 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}.

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.

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.

That progression always runs the same way, whatever the equation. 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.

So the claim “every equation of the form $x^2 + y^2 + Dx + Ey + F = 0$ is a circle” is simply false. If completing the square reaches $(x - 2)^2 + (y + 3)^2 = -7$ , the right side is negative: $\sqrt{-7}$ names no real radius, and the equation has no real 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. See it happen on one circle first, with three parallel lines.

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.

The pattern behind that picture is a single substitution, repeatable for any line and any circle. Substitute the line $y = mx + q$ into the origin-centered circle $x^2 + y^2 = r^2$ and collect powers of $x$ :

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

This substitution always produces a genuine quadratic in $x$ : its leading coefficient is $1 + m^2$ , which is at least $1$ and never zero. A quadratic has at most two real roots, and the line assigns exactly one $y$ to each $x$ , so distinct roots give distinct points. That is why a line can cross a circle at most twice, no matter where the circle is centered. A vertical line $x = a$ has no slope to substitute; put $x = a$ into the circle instead, which again leaves at most two real solutions for $y$ .

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 = (2mq)^2 - 4(1 + m^2)(q^2 - r^2) = 4\left[\, r^2(1 + m^2) - q^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.

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.

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.

Five outcomes for two circles, by the distance d between centers compared with the sum of the radii and the positive difference of the radii. Too far apart: d is greater than the sum, no shared points. Touching outside: d equals the sum, one point. Crossing: d is between the difference and the sum, two points. Touching inside: d equals the difference, one point. Nested: d is less than the difference, no shared points even though the circles are close. A sixth case, not pictured: equal centers and equal radii make the two circles identical, sharing every point.

That progression is controlled by a single number: compare the distance $d$ between centers with the sum $r_1 + r_2$ and the difference $|r_1 - r_2|$ . 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.

One case sits outside that pattern. If the two circles share the same center and the same radius, so $d = 0$ and $r_1 = r_2$ , they are not two circles but one, and every point on it is shared.

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.

Worked example 5 Do $x^2 + y^2 = 25$ and $(x - 8)^2 + y^2 = 16$ intersect?

Read off the centers and radii: the first circle has center $(0, 0)$ and radius $r_1 = 5$ ; the second has center $(8, 0)$ and radius $r_2 = 4$ .

The centers sit $d = 8$ apart along the $x$ -axis. Compare $d$ with the two bounds:

r_1 + r_2 = 9, \qquad |r_1 - r_2| = 1 .

Since $1 < 8 < 9$ , the distance falls strictly between the bounds, so the circles cross at two points.

Check your understanding

How many points do the circles $x^2 + y^2 = 36$ and $(x - 1)^2 + y^2 = 4$ share?

A two B one C none, they are too far apart D none, one circle is inside the other

When two distinct circles 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. If the centers differ, the remaining $x$ - and $y$ -terms do not both cancel too, so what is left is 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.

The general form has one more use worth knowing about. Because $D$ , $E$ , and $F$ each appear to the first power, an equation built from a known point $(x_0, y_0)$ is linear in those three unknowns. Three points, so long as they do not all lie on one straight line, give three linear equations and pin down exactly one circle.

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 wrong number as the radius. The right side of center-radius form is $r^2$ , not $r$ : a right side of $20$ means $r = \sqrt{20} = 2\sqrt{5}$ . And given the two endpoints of a diameter, the radius is half the distance between them, not the whole distance.
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 $1$ .
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.
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 $x^2 + y^2 + Dx + Ey + F = 0$ . Completing the square once, with letters, converts it back to $\left(x + \tfrac{D}{2}\right)^2 + \left(y + \tfrac{E}{2}\right)^2 = \tfrac{D^2 + E^2 - 4F}{4}$ : the center is always $\left(-\tfrac{D}{2}, -\tfrac{E}{2}\right)$ , and the sign of $D^2 + E^2 - 4F$ decides everything, positive for a circle of radius $\tfrac{1}{2}\sqrt{D^2 + E^2 - 4F}$ , zero for the single point $(h, k)$ , negative for 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. Distinct circles share no points when $d > r_1 + r_2$ or $d < |r_1 - r_2|$ (one inside the other), exactly one point when $d$ equals either bound, and two points strictly between them. Equal centers and equal radii describe the same circle, with every point shared.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Conic sections: introduction (Khan Academy)
Distance and midpoint formulas; circles (free textbook) (OpenStax)
Equation of a circle (Math is Fun)
Circles: center-radius form (Purplemath)

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Why a line meets a circle at most twice, from any center

A line meets a circle in at most two points#

Substituting $y = mx + q$ into $x^2 + y^2 = r^2$ collects into $(1 + m^2)x^2 + 2mq\,x + (q^2 - r^2) = 0$ . Its leading coefficient is $1 + m^2$ , and since $m^2 \geq 0$ for every real slope, $1 + m^2 \geq 1$ , 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 $y = mx + q$ . 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(q - k) - 2h \,\right] x + \left[\, h^2 + (q - 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 lines are tangent to a circle of a given slope

Worked example Which lines $y = x + q$ 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 + q)^2 = 25 \;\Longrightarrow\; 2x^2 + 2qx + (q^2 - 25) = 0 .

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

-4q^2 + 200 = 0 \;\Longrightarrow\; q^2 = 50 \;\Longrightarrow\; q = \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 $\Delta = 4[r^2(1 + m^2) - q^2]$ 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.

Finding the circle through three given points

The general form looks uglier than the center-radius form, but it earns its keep in one situation the tidy form cannot handle. Treat $D$ , $E$ , and $F$ in $x^2 + y^2 + Dx + Ey + F = 0$ as the unknowns and a known point $(x_0, y_0)$ as the data: this equation is linear in those unknowns, 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, so three points give three linear equations in three unknowns, a system you already know how to solve.

Worked example 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.

There is a second way to find that center, straight from the distance condition. Call the unknown center $(x, y)$ . It is as far from $(1, 1)$ as from $(2, 4)$ , so square that condition:

(x - 1)^2 + (y - 1)^2 = (x - 2)^2 + (y - 4)^2 .

The $x^2$ and $y^2$ cancel from both sides, leaving the line $x + 3y = 9$ . It passes through the midpoint $\left(\tfrac{3}{2}, \tfrac{5}{2}\right)$ of the two points, and its slope $-\tfrac{1}{3}$ is the negative reciprocal of the slope $3$ of the segment joining them: it is the perpendicular bisector of that segment. The same cancellation happens for any two different points, so the points equally far from two given points always form a line. Doing it again for $(2, 4)$ and $(5, 3)$ gives $3x - y = 7$ , and the two lines cross at $(3, 2)$ , the center found above.

Three points determine the circle, provided they do not all lie on one line: if they do, both bisectors are perpendicular to that line, so they are parallel and never meet, and no circle exists (in the algebra, the elimination ends in a contradiction). Four points are usually one demand too many for any single circle to satisfy.

In the figure below, $P$ , $Q$ and $R$ start at $(1, 1)$ , $(2, 4)$ and $(5, 3)$ . Drag $P$ , $Q$ and $R$ anywhere on the grid and watch the dashed bisectors meet at the center $C$ . Then line the three points up and watch the bisectors turn parallel, leaving no center and no circle.

Three points not on one line fix one circle

The points are P(1, 1), Q(2, 4) and R(5, 3). The perpendicular bisectors of PQ and QR cross at C(3, 2). CP = CQ = CR = √5, so the circle with center C and radius √5 passes through P, Q and R. (x - 3)² + (y - 2)² = 5.

A coordinate plane with three points P, Q and R, the segments PQ and QR, and the perpendicular bisector of each drawn dashed through its midpoint. When the three points are different and not on one line, the bisectors cross at the center C of the circle through all three. When three different points lie on one line, the bisectors are parallel and no circle passes through all three. When two of the points coincide, their segment and its bisector disappear.

A bit of history (optional)

Fix one point, fix one distance, and collect everything that far away. This lesson turned that rule into an equation. But is it the only rule that draws a circle?

Apollonius of Perga, a Greek geometer working around 200 BCE, found a second one, and it looks nothing like the first. He is the man who gave the ellipse, the parabola and the hyperbola the names they still carry. So he had reason to hunt for odd ways of describing a familiar curve. His rule uses two fixed points rather than one. Collect every point whose distance to the first is exactly twice its distance to the second.

Nothing in that description sounds round. There is no center named anywhere in it, and no radius either. The condition is a ratio rather than a length, and the two fixed points are not even treated alike. Yet the points that satisfy it form a perfect circle, and neither of the two fixed points sits at its center. Any positive ratio behaves the same way, with one exception. At a ratio of exactly $1$ the two distances are equal, and the set straightens out into a line: the perpendicular bisector you would draw between the points.

You can settle all of this yourself, with nothing beyond this lesson. Write both distances with the distance formula, square both sides, and gather the terms. The $x^2$ and the $y^2$ arrive with equal coefficients and no $xy$ term between them, which is exactly the fingerprint you learned to read.