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MathLabwith Mr. Ahmed
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1.9 The Coordinate Plane; Graphs of Equations; Circles

Read a circle’s center and radius.

Before you solve

Ideas in this lesson

Review the ideas below. Select one to see worked steps and five practice questions.

01Use distance and Pythagorean reasoningWorked steps · five questions02Find midpoints and recover endpointsWorked steps · five questions03Read, plot, and check graphs of equationsWorked steps · five questions04Find and interpret axis interceptsWorked steps · five questions05Test symmetry and transform coordinate figuresWorked steps · five questions06Build circle equations from geometric informationWorked steps · five questions07Complete squares and classify circle equationsWorked steps · five questions08Describe and graph coordinate regionsWorked steps · five questions09Write circle equations from diameter endpointsWorked steps · five questions10Use tangency to one or both coordinate axesWorked steps · five questions11Prove coordinate geometry and calculate areasWorked steps · five questions12Apply midpoints to triangles and quadrilateralsWorked steps · five questions
Lesson brief and rules

1.9 The Coordinate Plane; Graphs of Equations; Circles

The standard circle equation is (x−h)2+(y−k)2=r2(x - h)^{2} + (y - k)^{2} = r^{2}. Its center is (h,k)(h, k) and its radius is rr, where r>0r > 0.

Remember
The right side is the radius squared, not the radius.
Worked example 11
01 / A cable runs straight from (−2,2)(-2,2) to (4,5)(4,5). Coordinates are in meters. Find its exact length.
02 / The horizontal change is 4−(−2)=64-(-2)=6.
03 / The vertical change is 5−2=35-2=3.
04 / Use the Pythagorean theorem: d=62+32=45d=\sqrt{6^{2}+3^{2}}=\sqrt{45}.
05 / Distance is nonnegative; keep the radical exact rather than rounding.
Worked example 2 · Use distance and Pythagorean reasoning

A point P=(u,5)P=(u,5) lies 55 units from (−2,1)(-2,1). Find all possible values of uu.

Square the distance condition: (u+2)2+42=25(u+2)^{2}+4^{2}=25.
Subtract 1616 to obtain (u+2)2=9(u+2)^{2}=9.
Both signs are possible: u+2=±3u+2=\pm3.
These two points lie on opposite sides of the vertical line through the center.

Read the graphs

Find midpoints and recover endpoints

The two plotted points are AA and BB. Read their coordinates from the graph, find ABAB, and construct the midpoint of ABAB.
Two points for distance and midpointTwo marked points on labeled coordinate axes; read each coordinate before calculating.−22468−2246xy
· Read the two plotted coordinates. Let A be the left point and B the right point.

Read the points as (−2,5)(-2,5) and (8,−1)(8,-1). Their coordinate changes are 1010 and −6-6.
The Distance Formula gives AB=102+(−6)2=136=2 34AB=\sqrt{10^{2}+(-6)^{2}}=\sqrt{136}=2\,\sqrt{34}.
Average the corresponding coordinates: M=(−2+82,5+(−1)2)=(3,2)M=(\dfrac{-2+8}{2},\dfrac{5+(-1)}2)=(3,2). Plot this point halfway along the segment.

Answer: A=(−2,5),B=(8,−1)A=(-2,5), B=(8,-1); distance 2 342\,\sqrt{34}; midpoint (3,2)(3,2).

Read, plot, and check graphs of equations

Read the plotted graph. Which equation agrees with its vertex and the two other marked points? From the graph, determine whether it is symmetric about the xx-axis, yy-axis, or origin.
Coordinate graphA parabola opens upward. Its marked vertex V is at (-2,1); marked points A and B are at (-4,5) and (0,5).-8-8-6-6-4-4-2-222446688xyVAB
A parabola opens upward. Its marked vertex V is at (-2,1); marked points A and B are at (-4,5) and (0,5).

The lowest point is (−2,1)(-2,1), so the graph is shifted −2-2 horizontally and 11 vertically from y=x2y=x^{2}.
Each point two horizontal units from the vertex is four vertical units higher, agreeing with 22=42^{2}=4.
The positive square term gives the observed upward opening.
Organize exact values before sketching; the paired values also show symmetry about x=−2x=-2.

Exact point-plotting values for the displayed parabola
xxyy
−4-455
−3-322
−2-211
−1-122
0055

The plotted parabola is mirrored across x=−2x=-2, not either coordinate axis. For example, its vertex (−2,1)(-2,1) has no matching point (−2,−1)(-2,-1), (2,1)(2,1), or (2,−1)(2,-1), excluding x-axis, y-axis and origin symmetry.

Answer: y=(x+2)2+1y=(x+2)^{2}+1 None of these three symmetries holds; its reflection axis is x=−2x=-2.

Find and interpret axis intercepts

Read all axis intercepts from the supplied graph, then verify them algebraically using y=(x−2)(x+1)(x−4)y=(x-2)(x+1)(x-4).
Coordinate-plane graphThe graph corresponds to the stated function or supplied points.−2−112345−10−5510xy
Solid: curve 1 · Read the axis crossings; exact verification uses the stated equation.

Set y=0y=0 to find xx-intercepts; set x=0x=0 to find yy-intercepts.
(0−2)(0+1)(0−4)=8(0-2)(0+1)(0-4)=8.
Retain every real solution and check it in the original equation.

Answer: (−1,0),(2,0),(4,0),(0,8)(-1,0),(2,0),(4,0),(0,8)

Test symmetry and transform coordinate figures

Reflect the triangle with vertices (1,2)(1,2), (4,2)(4,2), (2,5)(2,5) across the yy-axis. Give all three image vertices. State the general reflection rule, and recover the preimage of (−5,4)(-5,4).
Coordinate-plane graphThe graph corresponds to the stated function or supplied points.−4−224−1123456xy
· Reflect the entire triangle across the y-axis.

Reflection in the yy-axis changes xx to its negative and leaves yy unchanged.
Apply the rule to every vertex, and join the image vertices in the original order.
The rule is (x,y)↦(−x,y)(x,y)\mapsto(-x,y) and is its own inverse. Therefore (−5,4)(-5,4) comes from (5,4)(5,4).

Answer: (−1,2),(−4,2),(−2,5)(-1,2),(-4,2),(-2,5); rule (x,y)↦(−x,y)(x,y)\mapsto(-x,y); preimage (5,4)(5,4).

Build circle equations from geometric information

Write the equation of the circle shown.
Coordinate graphA circle has marked center C at (-2,1) and rightmost point R at (1,1).-6-6-4-4-2-2224466xyCR
A circle has marked center C at (-2,1) and rightmost point R at (1,1).

Read the center (−2,1)(-2,1) from the graph.
The horizontal distance from CC to RR is 33, the radius.
Substitute into (x−h)2+(y−k)2=r2(x-h)^{2}+(y-k)^{2}=r^{2} and square the radius.

Answer: (x+2)2+(y−1)2=9(x+2)^{2}+(y-1)^{2}=9

Describe and graph coordinate regions

Which inequalities describe the shaded region, including its boundary conventions?
Coordinate graphShading is to the right of the solid vertical boundary x=1 and below the dashed horizontal boundary y=2; only their overlap is shaded.-6-6-4-4-2-2224466xy
Shading is to the right of the solid vertical boundary x=1 and below the dashed horizontal boundary y=2; only their overlap is shaded.

The shaded points are on or to the right of the vertical boundary, so x≥1x\ge1.
They lie below the horizontal boundary, which is dashed and excluded, so y<2y<2.
Only the overlap is shaded, requiring “and” rather than “or.”

Answer: x≥1 and y<2x\ge1\text{ and }y<2

15 questions covering every idea, with a worked answer for each.Start practice ↗
Textbook reference: Stewart, Redlin & Watson, Precalculus: Mathematics for Calculus, 7th ed. (Cengage, 2016). The explanations and practice here are original, organized by textbook section.

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