Geometry · G.GPE.A.1

Finding the center and radius

A circle’s equation has been multiplied out, and its center and radius have vanished into the numbers. How do you get them back?

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Equation of a circleGeometry

Where you are 15 in Circles

Geometry11

Circles15

▴ 10 above Angles made by chords, secants and tangents Chord theorems Segment lengths in circles Equation of a circle

Finding the center and radius

When you reach for it

The standard form of a circle, (x - h)^2 + (y - k)^2 = r^2, shows the center (h, k) and the radius r at a glance. Multiply it out and they disappear. Take the circle with center (-2, 5) and radius 3:

(x + 2)^2 + (y - 5)^2 = 9

Expanding both squares gives x^2 + 4x + 4 + y^2 - 10y + 25 = 9, and collecting everything on the left,

x^2 + y^2 + 4x - 10y + 20 = 0

This is the same circle, every point of it and no other, but nothing in the equation says (-2, 5) or 3. The 4 and the 25 that came out of the squares have been combined with the -9, and all three are now hidden inside the 20.

An equation in this shape, x^2 + y^2 + Dx + Ey + F = 0 for some numbers D, E and F, is called the general form of a circle. It is how circles usually arrive when they come out of a calculation rather than from a stated center: fitting a circle through three points, describing a set of points by a condition on their distances, or any working that multiplied brackets out.

Reach for this move whenever an equation has an x^2 and a y^2 with the same coefficient and no xy term, and you need the center and the radius: to draw the circle, to compare it with another circle, to test points against it, or to find out whether the equation describes a circle at all. The move is completing the square, done twice: once for the x terms and once for the y terms.

The procedure

Where the center went

Expand the standard form with letters and the hiding place shows itself. The equation (x - h)^2 + (y - k)^2 = r^2 becomes x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2, and with everything moved to the left,

\begin{aligned} &x^2 + y^2 - 2hx - 2ky \\ &\qquad + (h^2 + k^2 - r^2) = 0 \end{aligned}

Three things can be read from this. The squares x^2 and y^2 both have coefficient 1, and there is no term with x and y multiplied together. The coefficient of x is -2h: the center’s first coordinate is still there, doubled and with its sign changed, and the coefficient of y is -2k in the same way. And the constant term mixes h^2, k^2 and r^2 together, so the radius cannot be read from it on its own.

Completing the square undoes exactly this. In x^2 + bx it takes half the coefficient b and builds the square \displaystyle \left(x + \frac{b}{2}\right)^2. Here b = -2h, and half of it is -h, which rebuilds (x - h)^2. The same happens in y, and what is left on the right is r^2.

Completing the square in each variable

Work on x^2 + y^2 + 4x - 10y + 20 = 0 as though its origin were unknown.

Gather each variable’s terms and move the constant. The terms of a sum may be put in any order (the commutative property of addition) and grouped in any way (the associative property), so the two x terms can sit together and the two y terms together. Subtracting 20 from both sides sends the constant across:

(x^2 + 4x) + (y^2 - 10y) = -20

Each bracket now has the shape x^2 + bx that completing the square works on, with nothing else inside it.

Complete both squares, adding to both sides. For x: half of 4 is 2, and 2^2 = 4. For y: half of -10 is -5, and (-5)^2 = 25. Put each number into its own bracket, and because this is an equation, add both numbers to the right side as well. Adding the same number to both sides keeps the equation true for exactly the same points:

\begin{aligned} &(x^2 + 4x + 4) + (y^2 - 10y + 25) \\ &\qquad = -20 + 4 + 25 = 9 \end{aligned}

The two squares are completed separately because the x part and the y part do not touch. There is no xy term to tie them together, so each completing number depends only on its own variable’s coefficient.

Write each bracket as a square. Each bracket is now a perfect square trinomial, and the number inside the square is the half found a moment ago, not its square:

(x + 2)^2 + (y - 5)^2 = 9

Read off the center and radius. This is the standard form. The bracket x + 2 is x - (-2), so h = -2; the bracket y - 5 gives k = 5; and the right side is r^2 = 9, so r = 3. The circle has center (-2, 5) and radius 3.

Figure 1The circle x^2 + y^2 + 4x - 10y + 20 = 0. Completing the square in x and in y turns it into (x + 2)^2 + (y - 5)^2 = 9, which names this center and this radius.

The answer can be trusted for a simple reason. Every step either rewrote one side without changing its value (reordering, grouping, writing a perfect square trinomial as a square) or did the same thing to both sides (subtracting 20, adding 4, adding 25). Each of those steps can be undone, so the last equation is true for exactly the same points (x, y) as the first. They are two names for one circle. Expanding the final equation, as at the start of this lesson, gives the original back, and that is the natural check.

What the procedure always gives

Do the same steps on x^2 + y^2 + Dx + Ey + F = 0 with letters. Moving F across gives (x^2 + Dx) + (y^2 + Ey) = -F. The completing numbers are \displaystyle \frac{D^2}{4} and \displaystyle \frac{E^2}{4}, and adding both to both sides,

\begin{aligned} &\left(x + \frac{D}{2}\right)^2 + \left(y + \frac{E}{2}\right)^2 \\ &\qquad = \frac{D^2 + E^2 - 4F}{4} \end{aligned}

So the center is \displaystyle \left(-\frac{D}{2}, -\frac{E}{2}\right): each coordinate of the center is minus half the coefficient of its variable. That makes a quick check on any answer. The right side is worth working out each time rather than memorizing, since the steps also cope with an equation that arrives untidily arranged.

Circle, point or nothing

The number on the right, call it d, came out of arithmetic, and nothing forces it to be positive. The left side is a sum of two squares, and a square of a real number is never negative, so:

If d > 0, the equation is a circle with radius \displaystyle \sqrt{d}.
If d = 0, a sum of two squares that are never negative can only be 0 when both squares are 0. The only solution is the center itself, a single point.
If d < 0, no point at all satisfies the equation, since the left side is never negative.

In terms of the general form, the equation is a circle exactly when D^2 + E^2 - 4F > 0, that is when D^2 + E^2 > 4F. Completing the square answers not only “where is the circle” but “is there one”.

When the squares carry a coefficient

Completing the square needs x^2 and y^2 standing alone. When both carry the same coefficient, as in 3x^2 + 3y^2 - 12x + 6y - 9 = 0, divide both sides of the equation by it first. Dividing both sides by the same non-zero number does not change which points satisfy the equation, and 0 \div 3 is still 0, so this is the circle x^2 + y^2 - 4x + 2y - 3 = 0, which completes to (x - 2)^2 + (y + 1)^2 = 8. This is a move for equations; the one-variable version for expressions is in completing the square with a leading coefficient.

If the two coefficients are different, no division makes them both 1, and the equation is not a circle at all. Boundaries gives an example of what it can be instead.

Worked examples

Example 1

A circle through the origin

Find the center and radius of the circle x^2 + y^2 - 8x + 6y = 0.

Solution

There is no constant term, so there is nothing to move. Group the x terms and the y terms:

(x^2 - 8x) + (y^2 + 6y) = 0

Half of -8 is -4, and (-4)^2 = 16. Half of 6 is 3, and 3^2 = 9. Add both numbers to both sides, so the right side becomes 0 + 16 + 9 = 25:

\begin{aligned} (x^2 - 8x + 16) + (y^2 + 6y + 9) &= 25 \\ (x - 4)^2 + (y + 3)^2 &= 25 \end{aligned}

The center is (4, -3) and the radius is \displaystyle \sqrt{25} = 5.

Check. With no constant term, (0, 0) satisfies the original equation, so the origin is on the circle. Its distance from the center is \displaystyle \sqrt{4^2 + 3^2} = 5, which is the radius, as it must be.

Example 2

One variable with nothing to complete

Find the center and radius of x^2 + y^2 + 10y - 11 = 0.

Solution

There is no x term. That means b = 0 for x, the completing number is 0^2 = 0, and x^2 is already the square (x - 0)^2. Only y needs work. Move the -11 across by adding 11 to both sides:

x^2 + (y^2 + 10y) = 11

Half of 10 is 5, and 5^2 = 25:

\begin{aligned} x^2 + (y^2 + 10y + 25) &= 11 + 25 \\ x^2 + (y + 5)^2 &= 36 \end{aligned}

The center is (0, -5), on the y axis, and the radius is 6.

Check. The point (6, -5) is 6 to the right of the center, so it should be on the circle. In the original equation, 36 + 25 - 50 - 11 = 0.

Example 3

Odd coefficients make fractions

Find the center and radius of x^2 + y^2 + 3x - 5y + 2 = 0.

Solution

Group and move the constant:

(x^2 + 3x) + (y^2 - 5y) = -2

Half of 3 is \displaystyle \frac{3}{2}, whose square is \displaystyle \frac{9}{4}. Half of -5 is \displaystyle -\frac{5}{2}, whose square is \displaystyle \frac{25}{4}. Fractions are normal here and there is no reason to turn them into decimals. Add both to both sides. Writing -2 as \displaystyle -\frac{8}{4}, the right side becomes \displaystyle -\frac{8}{4} + \frac{9}{4} + \frac{25}{4} = \frac{26}{4} = \frac{13}{2}, and the left side is two squares:

\left(x + \frac{3}{2}\right)^2 + \left(y - \frac{5}{2}\right)^2 = \frac{13}{2}

The center is \displaystyle \left(-\frac{3}{2}, \frac{5}{2}\right), which agrees with “minus half of each coefficient”. The radius is

r = \sqrt{\frac{13}{2}} = \frac{\sqrt{13}}{\sqrt{2}} = \frac{\sqrt{26}}{2} \approx 2.55

The right side is r^2, not r, so the square root is still needed after the squares are completed.

Example 4

A common coefficient and terms on both sides

Find the center and radius of the circle 4x^2 + 4y^2 + 8x + 31 = 24y.

Solution

First bring every term to one side, by subtracting 24y from both sides:

4x^2 + 4y^2 + 8x - 24y + 31 = 0

Both squares carry the coefficient 4, so divide both sides by 4. That leaves the same points, and the constant becomes a fraction:

x^2 + y^2 + 2x - 6y + \frac{31}{4} = 0

Group, and subtract \displaystyle \frac{31}{4} from both sides:

(x^2 + 2x) + (y^2 - 6y) = -\frac{31}{4}

Half of 2 is 1, and 1^2 = 1. Half of -6 is -3, and (-3)^2 = 9. On the right, write 1 + 9 = 10 as \displaystyle \frac{40}{4}:

\begin{aligned} (x + 1)^2 + (y - 3)^2 &= -\frac{31}{4} + \frac{40}{4} \\ &= \frac{9}{4} \end{aligned}

The center is (-1, 3) and the radius is \displaystyle \sqrt{\frac{9}{4}} = \frac{3}{2}.

Check in the equation exactly as it was given. The point \displaystyle \left(\frac{1}{2}, 3\right) is \displaystyle \frac{3}{2} to the right of the center. On the left, \displaystyle 4 \cdot \frac{1}{4} + 4 \cdot 9 + 8 \cdot \frac{1}{2} + 31 = 1 + 36 + 4 + 31 = 72, and on the right, 24 \cdot 3 = 72.

Had the 4 not been divided out, halving 8 would have given a completing number of 16 for 4x^2 + 8x, and 4x^2 + 8x + 16 is not a perfect square. The halving rule belongs to x^2 + bx.

Example 5

A point, and nothing

Decide what each equation describes.

(a) x^2 + y^2 - 4x + 2y + 5 = 0 \qquad (b) x^2 + y^2 + 6x - 8y + 30 = 0

Solution

Both look like circles. Completing the square decides.

(a) Group and move the constant: (x^2 - 4x) + (y^2 + 2y) = -5. Add 4 and 1 to both sides:

(x - 2)^2 + (y + 1)^2 = 0

Two squares that are never negative add to 0 only when both are 0, so x = 2 and y = -1. The equation describes the single point (2, -1), a circle shrunk to radius 0.

(b) Group and move the constant: (x^2 + 6x) + (y^2 - 8y) = -30. Add 9 and 16 to both sides:

(x + 3)^2 + (y - 4)^2 = -5

The left side is never negative, so it is never -5. No point satisfies this equation, and its graph is empty. By the test in the procedure, D^2 + E^2 = 36 + 64 = 100 is less than 4F = 120, which says the same thing.

Example 6

A family of circles

For which numbers k is x^2 + y^2 - 2x + 4y + k = 0 a circle? For which k is its radius 2?

Solution

Complete the squares with k left as a letter. Move it across, then add 1 and 4:

\begin{aligned} (x^2 - 2x) + (y^2 + 4y) &= -k \\ (x - 1)^2 + (y + 2)^2 &= 5 - k \end{aligned}

The right side is the d of the procedure. It is a circle exactly when 5 - k > 0, that is when k < 5. When k = 5 it is the single point (1, -2), and when k > 5 it is nothing.

The radius is 2 when 5 - k = 2^2 = 4, so k = 1.

Every circle in this family has the same center, (1, -2), because the center depends only on the coefficients of x and y. Changing the constant changes only the radius. Circles that share a center are called concentric.

Example 7

Two circles that touch

Show that the circles x^2 + y^2 - 4x - 6y + 9 = 0 and x^2 + y^2 + 8x + 10y - 23 = 0 meet at exactly one point, and find it.

Solution

Neither equation shows anything until it is in standard form, so complete the squares in both.

For the first, (x^2 - 4x) + (y^2 - 6y) = -9; adding 4 and 9 gives (x - 2)^2 + (y - 3)^2 = 4. Center C_1 = (2, 3), radius 2.

For the second, (x^2 + 8x) + (y^2 + 10y) = 23; adding 16 and 25 gives (x + 4)^2 + (y + 5)^2 = 64. Center C_2 = (-4, -5), radius 8.

The The distance formula gives the distance between the centers:

\begin{aligned} C_1C_2 &= \sqrt{(2 - (-4))^2 + (3 - (-5))^2} \\ &= \sqrt{36 + 64} = 10 \end{aligned}

and 10 = 2 + 8, the sum of the radii. That forces the circles to touch. If a point P is on both circles, then PC_1 = 2 and PC_2 = 8, so C_1P + PC_2 = 10 = C_1C_2. By the Triangle inequality, C_1C_2 \le C_1P + PC_2, with equality only when P lies on the segment C_1C_2. So any common point is on that segment, 2 units from C_1, and there is exactly one such point.

Figure 2The two circles of Example 7. The centers are 10 apart, which is 2 + 8, so the circles meet only at the point \displaystyle T = \left(\frac{4}{5}, \frac{7}{5}\right) on the segment joining the centers.

To find it, go from C_1 toward C_2, which is a step of (-6, -8), and stop \displaystyle \frac{2}{10} of the way along:

\begin{aligned} T &= \left(2 + \tfrac{2}{10}(-6),\ 3 + \tfrac{2}{10}(-8)\right) \\ &= \left(\frac{4}{5}, \frac{7}{5}\right) \end{aligned}

Check that T is on both circles. For the first, \displaystyle \left(\frac{4}{5} - 2\right)^2 + \left(\frac{7}{5} - 3\right)^2 = \frac{36}{25} + \frac{64}{25} = 4. For the second, \displaystyle \left(\frac{4}{5} + 4\right)^2 + \left(\frac{7}{5} + 5\right)^2 = \frac{576}{25} + \frac{1024}{25} = 64.

None of this could be seen in the general forms. The comparison of the distance with the radii is only possible once the centers and radii are known.

Boundaries

The coefficients of x^2 and y^2 must be equal, and there must be no xy term. Both are true of every circle, as the expansion of the standard form shows. The equation x^2 + 4y^2 = 4 is not a circle: the points (2, 0) and (0, 1) both satisfy it, and they are at different distances, 2 and 1, from the origin. Its graph is an ellipse. An xy term can make an equation that is not a circle look like one, as Exercise 25 shows.

Divide by a common coefficient only in an equation. Dividing both sides by the same non-zero number keeps the same points. If that coefficient is 0 there are no squares left, and the equation is a line (or, when the x and y terms vanish too, nothing or the whole plane).

Both completing numbers go on both sides. Each is added to the left, so each must be added to the right as well, or the equation changes.

The right side decides what the equation is. Positive gives a circle of radius \displaystyle \sqrt{d}, zero gives a single point, negative gives no points. An equation in general form is not a circle until this has been checked.

The right side is r^2. The radius is its positive square root, and it may be a fraction or a surd, as in Example 3.

This is a circle in the plane. With three variables, x^2 + y^2 + z^2 + Dx + Ey + Fz + G = 0 is completed three times, once per variable, and the result is the equation of a sphere.

Where this goes next

Once a circle is in standard form, everything in the equation of a circle applies to it: testing whether a point is inside, on or outside, and finding where a line meets it. With the center known, a tangent can be drawn at any point of the circle, since a tangent is perpendicular to the radius there. The circle through three given points is found by putting each point into the general form, solving for D, E and F, and then completing the square; its center is the circumcenter of the triangle the three points make (Exercise 19).

Completing the square to find a center is the same move that vertex form makes on a parabola, where one variable is squared instead of two. When the coefficients of x^2 and y^2 differ, completing the square in each variable still works, and the shape that comes out is an ellipse or a hyperbola. Sorting a second-degree equation into its shape this way is how a conic is identified from its equation, and the circle is the first of the Conic sections.

Exercises

Find the center and radius

x^2 + y^2 + 6x - 2y - 15 = 0
x^2 + y^2 - 10x + 16 = 0
x^2 + y^2 - 4x - 12y + 4 = 0
x^2 + y^2 + 8x + 8y + 14 = 0
x^2 + y^2 - 2y = 0

Fractions, coefficients and untidy equations

Find the center and radius.

\displaystyle x^2 + y^2 + x - 3y - \frac{3}{2} = 0
x^2 + y^2 - 5x + 3y = 0
3x^2 + 3y^2 - 18x + 12y + 27 = 0
4x^2 + 4y^2 - 4x + 8y - 11 = 0
x^2 + y^2 = 6x - 10y + 2
(x + 1)^2 + y^2 = 4x + 2y + 7

Circle, point or nothing

Decide whether each equation describes a circle, a single point or no points, and give the center and radius of any circle.

x^2 + y^2 - 6x + 4y + 13 = 0
x^2 + y^2 + 2x - 2y + 6 = 0
x^2 + y^2 - 8x + 2y + 8 = 0
Complete the square in x and in y for x^2 + 3y^2 - 4x + 6y = 0, and explain why the result is not the equation of a circle.
For which numbers k is x^2 + y^2 + 6x - 4y + k = 0 a circle? For which k is its radius 3?
For which numbers c is x^2 + y^2 - 2cx + 4y + 8 = 0 a circle? For which c is its radius \displaystyle \sqrt{5}?

Using the center and radius

Is the point (1, 2) inside, on or outside the circle x^2 + y^2 - 6x + 2y - 15 = 0?
A circle has general form x^2 + y^2 + Dx + Ey + F = 0 and passes through (0, 0), (10, 0) and (0, 4). Find D, E and F, then the center and radius.
Find the centers and radii of x^2 + y^2 - 2x - 4y - 4 = 0 and x^2 + y^2 - 14x - 20y + 145 = 0, and decide whether the two circles meet.
Find the area of the disc bounded by the circle x^2 + y^2 + 4x - 6y - 3 = 0.
Find the point of the circle x^2 + y^2 - 12x - 16y + 75 = 0 nearest the origin, and its distance from the origin.

Reasoning

Show that expanding (x - h)^2 + (y - k)^2 = r^2 gives an equation of the form x^2 + y^2 + Dx + Ey + F = 0, and write D, E and F in terms of h, k and r.
Explain why x^2 + y^2 + Dx + Ey + F = 0 is a circle exactly when D^2 + E^2 > 4F, and say what it describes when D^2 + E^2 = 4F and when D^2 + E^2 < 4F.
The equation x^2 + y^2 + 2xy - 4 = 0 has equal coefficients on x^2 and y^2. Show that it is not a circle by finding what it does describe.
When completing the square in x and y together, both completing numbers are added to the right side. Explain why adding only one of them, or neither, would give the wrong circle.

Answers

Center (-3, 1), radius 5. (x^2 + 6x) + (y^2 - 2y) = 15; add 9 and 1: (x + 3)^2 + (y - 1)^2 = 25.
Center (5, 0), radius 3. (x^2 - 10x) + y^2 = -16; add 25: (x - 5)^2 + y^2 = 9.
Center (2, 6), radius 6. (x^2 - 4x) + (y^2 - 12y) = -4; add 4 and 36: (x - 2)^2 + (y - 6)^2 = 36.
Center (-4, -4), radius \displaystyle 3\sqrt{2}. (x^2 + 8x) + (y^2 + 8y) = -14; add 16 and 16: (x + 4)^2 + (y + 4)^2 = 18, and \displaystyle \sqrt{18} = 3\sqrt{2}.
Center (0, 1), radius 1. Add 1 to both sides: x^2 + (y - 1)^2 = 1.
Center \displaystyle \left(-\frac{1}{2}, \frac{3}{2}\right), radius 2. Add \displaystyle \frac{1}{4} and \displaystyle \frac{9}{4}: \displaystyle \left(x + \frac{1}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{3}{2} + \frac{10}{4} = 4.
Center \displaystyle \left(\frac{5}{2}, -\frac{3}{2}\right), radius \displaystyle \frac{\sqrt{34}}{2}. Add \displaystyle \frac{25}{4} and \displaystyle \frac{9}{4}: \displaystyle \left(x - \frac{5}{2}\right)^2 + \left(y + \frac{3}{2}\right)^2 = \frac{17}{2}, and \displaystyle \sqrt{\frac{17}{2}} = \frac{\sqrt{34}}{2}.
Center (3, -2), radius 2. Divide by 3: x^2 + y^2 - 6x + 4y + 9 = 0. Then (x^2 - 6x) + (y^2 + 4y) = -9; add 9 and 4: (x - 3)^2 + (y + 2)^2 = 4.
Center \displaystyle \left(\frac{1}{2}, -1\right), radius 2. Divide by 4: \displaystyle (x^2 - x) + (y^2 + 2y) = \frac{11}{4}; add \displaystyle \frac{1}{4} and 1: \displaystyle \left(x - \frac{1}{2}\right)^2 + (y + 1)^2 = 4.
Center (3, -5), radius 6. Gather: (x^2 - 6x) + (y^2 + 10y) = 2; add 9 and 25: (x - 3)^2 + (y + 5)^2 = 36.
Center (1, 1), radius \displaystyle 2\sqrt{2}. The bracket is not the final form, so expand it: x^2 + 2x + 1 + y^2 = 4x + 2y + 7, so (x^2 - 2x) + (y^2 - 2y) = 6. Add 1 and 1: (x - 1)^2 + (y - 1)^2 = 8.
A single point, (3, -2). Add 9 and 4 to -13: (x - 3)^2 + (y + 2)^2 = 0.
No points. Add 1 and 1 to -6: (x + 1)^2 + (y - 1)^2 = -4, and the left side is never negative.
A circle with center (4, -1) and radius 3. Add 16 and 1 to -8: (x - 4)^2 + (y + 1)^2 = 9.
(x - 2)^2 + 3(y + 1)^2 = 7. In y, take out the 3 first: 3(y^2 + 2y), and completing adds 3 \cdot 1 = 3 to the right, with 4 from x. The y square still carries a 3 that the x square does not, so the points are not all at one distance from (2, -1): \displaystyle (2 + \sqrt{7}, -1) is \displaystyle \sqrt{7} away and \displaystyle \left(2, -1 + \sqrt{\frac{7}{3}}\right) is only \displaystyle \sqrt{\frac{7}{3}} away. It is an ellipse.
A circle when k < 13; radius 3 when k = 4. Add 9 and 4: (x + 3)^2 + (y - 2)^2 = 13 - k, which must be positive, and 13 - k = 9 gives k = 4.
A circle when c < -2 or c > 2; radius \displaystyle \sqrt{5} when c = 3 or c = -3. Add c^2 and 4: (x - c)^2 + (y + 2)^2 = c^2 - 4. This is positive exactly when c^2 > 4, and c^2 - 4 = 5 gives c^2 = 9.
Inside. The circle is (x - 3)^2 + (y + 1)^2 = 25, and (1 - 3)^2 + (2 + 1)^2 = 13, which is less than 25. Putting (1, 2) straight into the general form gives 1 + 4 - 6 + 4 - 15 = -12, which is 13 - 25: the left side of the general form is always the squared distance to the center minus r^2.
D = -10, E = -4, F = 0; center (5, 2), radius \displaystyle \sqrt{29}. The origin gives F = 0; then (10, 0) gives 100 + 10D = 0 and (0, 4) gives 16 + 4E = 0. So x^2 + y^2 - 10x - 4y = 0, and adding 25 and 4 gives (x - 5)^2 + (y - 2)^2 = 29. The center is the midpoint of (10, 0) and (0, 4).
Centers (1, 2) and (7, 10), radii 3 and 2; the circles do not meet. The equations complete to (x - 1)^2 + (y - 2)^2 = 9 and (x - 7)^2 + (y - 10)^2 = 4. The centers are \displaystyle \sqrt{36 + 64} = 10 apart. A common point P would have PC_1 + PC_2 = 3 + 2 = 5, less than 10, which the triangle inequality forbids.
16\pi. Add 4 and 9 to 3: (x + 2)^2 + (y - 3)^2 = 16, so r = 4 and the area is \pi r^2 = 16\pi.
The point (3, 4), at distance 5. The circle is (x - 6)^2 + (y - 8)^2 = 25, center C = (6, 8), radius 5, and \displaystyle OC = \sqrt{36 + 64} = 10. For any point P on the circle, OC \le OP + PC by the triangle inequality, so OP \ge 10 - 5 = 5, with equality only on the segment OC. Halfway along it is (3, 4), and 9 + 16 - 36 - 64 + 75 = 0 puts it on the circle.
D = -2h, E = -2k, F = h^2 + k^2 - r^2. Expanding gives x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2, and subtracting r^2 from both sides and reordering gives x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0.
Completing the square gives \displaystyle \left(x + \frac{D}{2}\right)^2 + \left(y + \frac{E}{2}\right)^2 = \frac{D^2 + E^2 - 4F}{4}. The left side is a sum of two squares, never negative, and it takes every value from 0 upwards. If the right side is positive the solutions are the points at distance \displaystyle \frac{\sqrt{D^2 + E^2 - 4F}}{2} from \displaystyle \left(-\frac{D}{2}, -\frac{E}{2}\right), a circle. If it is 0 only that center point satisfies the equation, and if it is negative no point does.
Two parallel lines, x + y = 2 and x + y = -2. The first three terms are a perfect square, x^2 + 2xy + y^2 = (x + y)^2, so the equation is (x + y)^2 = 4, which holds exactly when x + y = 2 or x + y = -2. The xy term is what stops it being a circle.
Adding a number to one side only changes the equation, so it is no longer true for the same points. Adding both numbers to the left and both to the right adds the same total to each side, which keeps exactly the original points. With only one added on the right, the right side is too small by the other, and the result is a circle with the right center and the wrong radius (or no circle at all), not the circle the equation described.