Coordinate geometry

Learning objectives

Connect distances, midpoints, slopes, and perpendicularity to geometric figures; write and interpret circle equations; complete the square; analyze translations, radii, tangency, and intersections; use algebra to confirm what a graph suggests.

Distance, midpoint, and direction

For points (x₁, y₁) and (x₂, y₂), the horizontal and vertical displacements form the legs of a right triangle. The Pythagorean theorem gives the distance formula:

$\sqrt{(x_{2} − x_{1})^{2} + (y_{2} − y_{1})^{2}}$

The midpoint averages the corresponding coordinates:

$\frac{x_{1} + x_{2}}{2}$

If a midpoint and one endpoint are known, work backward: $x 2 = 2x M - x 1$ and $y 2 = 2y M - y 1 .$

For a nonvertical line, its slope is $m = (y 2 - y 1)/(x 2 - x 1).$ Distinct parallel nonvertical lines have equal slopes. Perpendicular lines with defined, nonzero slopes have slopes whose product is −1: they are negative reciprocals. Horizontal and vertical lines are perpendicular to each other, but a vertical line’s slope is undefined, so do not force them into the product formula.

A line through (x₀, y₀) with slope m has equation $y - y 0 = m(x - x 0).$ A perpendicular bisector passes through a segment’s midpoint at a right angle to the segment. Points on that line are equidistant from the segment’s endpoints, a useful way to locate a circle’s center.

A circle equation is a distance statement

A circle with center (h, k) and positive radius r has equation

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

This simply says that the squared distance from (x, y) to the fixed center is r². Every point on the circle satisfies the equation. Conversely, every real point satisfying the equation lies on the circle.

For a point above and to the right of the center, the signed coordinate differences are also the positive leg lengths shown. Squaring makes the equation valid in every direction.

Read signs and squares carefully

The center of $(x + 2) 2 + (y - 5) 2 = 49$ is (−2, 5) because $x + 2 = x - (-2).$ The radius is 7, not 49. A quick check is to substitute the center: both squared terms should become zero.

For center (h, k) and radius r, the extreme coordinates are

$x min = h - r,$ $x max = h + r,$ $y min = k - r,$ $y max = k + r.$

To classify a point, compare its squared distance to r²: less means inside the disk, equal means on the circle, and greater means outside. No square roots are needed for that comparison.

Completing the square reveals center and radius

Expanded equations hide the center. The key identity is

$\frac{D}{2}$

Halve the linear coefficient, then square it. Do this separately for the x terms and the y terms. If you add a number to one side of the equation, add the same number to the other side. For the normalized equation $x 2 + y 2 + Dx + Ey + F = 0,$

$\frac{D}{2}$

Thus the center is $(-D/2, -E/2),$ and the right side is r². Use this as a summary of completing the square, not as a formula to apply before normalizing the equation.

Normalize first; then check that a circle exists If the coefficients of x² and y² are the same nonzero number, divide the entire equation by that number first. In the resulting standard form, a positive right side gives a circle; zero gives only its center point; a negative right side gives no real points. Unequal squared-term coefficients do not give this standard circle form merely by completing the square.

Transformations, tangency, and intersections

Translating a circle right by a and up by b changes its center from (h, k) to $(h + a, k + b)$ without changing its radius. Multiplying the radius by q multiplies the right side r² by q². A dilation about the origin also multiplies the center coordinates by q; an enlargement about the circle’s own center does not move the center. Always identify the transformation’s fixed point.

A circle tangent to the x-axis has radius |k|; one tangent to the y-axis has radius |h|. More generally, the radius equals the perpendicular distance from the center to the tangent line.

To find a line-circle intersection, substitute the line’s equation into the circle equation. The resulting quadratic typically has two, one, or no real solutions, matching a secant, tangent, or disjoint line. For a quadratic $Au 2 + Bu + C = 0$ with $A \neq 0,$ the discriminant is $B 2 - 4AC:$ positive gives two distinct real solutions, zero gives one repeated real solution, and negative gives none.

A graph is a useful check on location and the number of intersections. An exact algebraic calculation is still preferable when the answer is a radical or when a parameter must produce exact tangency. Use equal axis scales when visually judging distances or whether a graph looks circular.

15 worked examples

5.01. Distance between two points

Find the distance between (−3, 2) and (5, 8).

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Recognize the structure. Use the horizontal change 8 and vertical change 6 as right-triangle legs.

Work it through. The distance is

$\sqrt{(5 − (− {3))}^{2} + (8 − {2)}^{2}}$

Answer: 10 units.

Avoid the trap. Subtract coordinates of the same type: x from x and y from y. Also keep the double negative in $5 - (-3).$ The order of both point differences may be reversed because the differences are squared.

5.02. Recover an endpoint from a midpoint

The midpoint of segment AB is (1, 3). If A = (−6, 9), what are the coordinates of B?

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Recognize the structure. The midpoint averages the endpoint coordinates, so reverse that averaging process.

Work it through. Let $B = (u, v).$ Then

$\frac{−6 + u}{2}$ $\frac{9 + v}{2}$

Answer: (8, −3).

Key takeaway. The displacement from A to the midpoint is (7, −6). Moving that same displacement again reaches (8, −3), providing a geometric check.

5.03. A perpendicular line through a specified point

Line l passes through (2, −1) and (6, 7). Find an equation of the line perpendicular to l that passes through (4, 3).

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Recognize the structure. Find the original slope, take its negative reciprocal, then use the specified point.

Work it through. The slope of l is $(7 - (-1))/(6 - 2) = 8/4 = 2.$ The perpendicular slope is −1/2. Using point-slope form, $\frac{1}{2}$

Answer: $\frac{1}{2}$

Avoid the trap. A negative reciprocal is −1/2, not −2. Check both the slope product, $2(-1/2) = -1,$ and that (4, 3) satisfies the final equation.

5.04. Write a circle equation from center and radius

A circle has center (−2, 5) and radius 7. Write its equation.

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Recognize the structure. Substitute the signed center coordinates into $(x - h) 2 + (y - k) 2 = r 2 .$

Work it through. Use $h = -2,$ $k = 5,$ and $r = 7:$

$(x - (-2)) 2 + (y - 5) 2 = 7 2 .$

Answer: $(x + 2) 2 + (y - 5) 2 = 49.$

Key takeaway. The point (−2, 12) is 7 units directly above the center; substituting it gives $0 + 49 = 49.$ Checking a simple point helps catch both a center-sign error and a radius-square error.

5.05. Read the rightmost point, not just the radius

A circle is represented by (x − 4)² + (y + 3)² = 20. What is the largest possible x-coordinate of a point on the circle? A) 4 + 20 B) 4 + 2 $\sqrt{5}$ C) 2 $\sqrt{5}$ D) −4 + 2 $\sqrt{5}$

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Recognize the structure. The rightmost point is one radius to the right of the center.

Work it through. The center is (4, −3) and the radius is $\sqrt{20}$ Therefore $\sqrt{5}$

Answer: B, $\sqrt{5}$ Another efficient route. Since $(y + 3) 2 \geq 0,$ the largest possible (x − 4) occurs when $y = -3,$ giving $(x - 4) 2 = 20$ and the positive branch $\sqrt{20}$

5.06. A point on the circle gives the radius squared

A circle is centered at (3, −2) and passes through (9, 6). Write its equation.

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Recognize the structure. The radius is the distance from the center to the given point, but the equation needs only its square.

Work it through. Compute

$r 2 = (9 - 3) 2 + (6 - (-2)) 2 = 6 2 + 8 2 = 100.$

Answer: $(x - 3) 2 + (y + 2) 2 = 100.$

Key takeaway. There is no need to take a square root and then square again. The radius is 10, but using the squared distance directly is more efficient.

5.07. A diameter gives both center and radius

A diameter of a circle has endpoints (−4, 1) and (8, 7). Write the circle’s equation.

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Recognize the structure. The center is the midpoint of the diameter, not either endpoint.

Work it through. The midpoint is $\frac{−4 + 8}{2}$

Using either endpoint, $r 2 = (8 - 2) 2 + (7 - 4) 2 = 36 + 9 = 45.$

Answer: $(x - 2) 2 + (y - 4) 2 = 45.$

Avoid the trap. The endpoint-to-endpoint squared distance is 180. That is $d 2 = 4r 2,$ so it must be divided by 4, not by 2, to obtain r².

5.08. Complete the square in both variables

A circle has equation x² + y² − 10x + 6y − 15 = 0. Find its center and radius.

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Recognize the structure. Group the variable terms, move the constant, and add the two completing-square amounts to both sides.

Work it through. Rearrange and complete:

$x 2 - 10x + y 2 + 6y = 15,$ $(x 2 - 10x + 25) + (y 2 + 6y + 9) = 15 + 25 + 9,$ $(x - 5) 2 + (y + 3) 2 = 49.$

Answer: Center (5, −3); radius 7.

Avoid the trap. The added amounts are $(-10/2) 2 = 25$ and $(6/2) 2 = 9,$ not 100 and 36. Add them to the right side as well.

5.09. Normalize before completing the square

A circle has equation 2x² + 2y² + 12x − 8y − 24 = 0. Find its radius.

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Recognize the structure. The common coefficient of the squared terms is 2, so divide the entire equation by 2 first.

Work it through. After dividing, $x 2 + y 2 + 6x - 4y - 12 = 0.$

Then $(x 2 + 6x + 9) + (y 2 - 4y + 4) = 12 + 9 + 4,$ so $(x + 3) 2 + (y - 2) 2 = 25.$

Answer: 5 units.

Avoid the trap. Halving only some coefficients changes the equation. Normalization must include the constant term as well as both linear terms.

5.10. Translate the center and enlarge the radius

A circle has equation (x + 1)² + (y − 4)² = 9. Its center is translated 5 units right and 2 units down. A new circle at that translated center has twice the original radius. Write the new circle’s equation.

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Recognize the structure. Treat center movement and radius scaling as separate operations.

Work it through. The original center is (−1, 4) and its radius is 3. The new center is $(-1 + 5, 4 - 2) = (4, 2),$ and the new radius is 6. Therefore the right side is $62 = 36.$

Answer: $(x - 4) 2 + (y - 2) 2 = 36.$

Avoid the trap. Doubling the radius multiplies r² by 4, not by 2. Also, moving right makes the center’s x-coordinate larger even though the equation contains $x - 4.$

5.11. Tangency to an axis determines a constant

The equation x² + y² − 8x + 4y + c = 0 represents a circle tangent to the x-axis. What is c?

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Recognize the structure. Complete the square to identify the center; its distance from the x-axis must equal the radius.

Work it through. Rewriting gives $(x - 4) 2 + (y + 2) 2 = 20 - c.$ The center is (4, −2), whose distance from the x-axis is 2. Tangency requires $r = 2,$ so $r 2 = 4 = 20 - c,$ yielding $c = 16.$

Answer: 16.

Key takeaway. The tangent point is (4, 0). Substituting it into the original equation with $c = 16$ gives $16 - 32 + 16 = 0,$ an independent check.

5.12. A horizontal line cuts a circle

The line y = 3 intersects the circle (x − 2)² + (y + 1)² = 25 at two points. What is the distance between those points?

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Recognize the structure. Substitute the common y-coordinate; then the distance is the difference of the two x-coordinates.

Work it through. At $y = 3,$ $(x - 2) 2 + 16 = 25 (x - 2) 2 = 9.$

Thus $x = 2 \pm 3,$ giving points (−1, 3) and (5, 3). Their horizontal separation is $5 - (-1) = 6.$

Answer: 6 units.

Another efficient route. The center-to-chord distance is 4 and the radius is 5. Half the chord is $\sqrt{25 − 16}$ so the full chord is 6.

5.13. Exact tangency through a discriminant

For a positive constant b, line y = 3x + b is tangent to the circle x² + y² = 25. What is b?

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Recognize the structure. Tangency means the substituted quadratic has exactly one real solution.

Work it through. Substitute $y = 3x + b:$

$x 2 + (3x + b) 2 = 25 10x 2 + 6bx + (b 2 - 25) = 0.$

Set the discriminant to zero:

$(6b) 2 - 4(10)(b 2 - 25) = 0 36b 2 - 40b 2 + 1000 = 0.$

Therefore $b 2 = 250.$ Since $b > 0,$ $\sqrt{10}$

Answer: $\sqrt{10}$

Avoid the trap. A graph that looks tangent does not establish the exact parameter. Also, b is a y-intercept, not the radius; the line’s perpendicular distance from the origin is smaller than b.

5.14. An inscribed rectangle from circle coordinates

A rectangle has vertices (a, 6), (−a, 6), (−a, −6), and (a, −6), where a > 0. All four vertices lie on x² + y² = 100. What is the rectangle’s area?

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Recognize the structure. One vertex determines a; symmetry gives the full width and height.

Work it through. Substitute (a, 6) into the circle equation:

$a 2 + 36 = 100 a 2 = 64 a = 8.$

The width is $8 - (-8) = 16,$ and the height is $6 - (-6) = 12.$ The area is $16(12) = 192.$

Answer: 192 square units.

Avoid the trap. a and 6 are half-dimensions. Multiplying 8 by 6 gives only one quarter of the rectangle’s area.

5.15. Find a center using a perpendicular bisector

A circle passes through A = (1, 2) and B = (5, 2). Its center lies on the line y = 2x − 1. Write the circle’s equation.

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Recognize the structure. A center is equidistant from A and B, so it lies on the perpendicular bisector of chord AB.

Work it through. The chord is horizontal and has midpoint (3, 2), so its perpendicular bisector is $x = 3.$ At the intersection with $y = 2x - 1,$ the center is (3, 5). Therefore

$r 2 = (1 - 3) 2 + (2 - 5) 2 = 4 + 9 = 13.$

Answer: $(x - 3) 2 + (y - 5) 2 = 13.$

Key takeaway. Two points alone do not determine a unique circle. The added center-line condition selects a unique center here because it intersects the perpendicular bisector at one point.

Mastery check and error prevention

Translate standard form into center and radius without hesitation, then reverse the process. Complete the square in expanded equations while keeping both sides balanced. Explain why a tangent gives one intersection, and distinguish a circle’s radius from its squared radius or diameter.

A compact graphing check

For Example 5.12, graph the circle and the horizontal line as two equations, then compare the intersection coordinates with (−1, 3) and (5, 3). The algebra establishes the exact result; the graph checks that both points, the horizontal chord, and the center are in the expected positions. Do not infer an exact radical from a rounded coordinate display.