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The Equation of a Circle: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two possible stops

The figure marks a starting point $A$ and possible stops $B$ and $C$ . Which stop is closer to $A$ by straight-line distance?

The starting point $A$ and the two possible stops $B$ and $C$ .

Text description of this figure

A square coordinate grid. The horizontal x-axis runs from negative four to three and the vertical y-axis runs from negative two to five, with tick marks, number labels and gridlines at every whole number, equal unit lengths on both axes, and the origin labeled 0. Three points are plotted, each marked with a filled dot and labeled with its letter only: A sits three units left of the y-axis and one unit above the x-axis; B sits one unit right of the y-axis and four units above the x-axis; C sits two units right of the y-axis and one unit below the x-axis. One thin straight segment joins A to B and another thin straight segment joins A to C. No lengths, right triangles or other guide lines are drawn.
1. Hint 1
  Compare the two distances using their horizontal and vertical gaps.
2. Hint 2
  Comparing squared distances is enough to identify the shorter distance.
Answer

$B$ is closer to $A$ .

Full solution

The points read from the grid are $A=(-3,1)$ , $B=(1,4)$ , and $C=(2,-1)$ .
The squared distance to $B$ is
$AB^2=4^2+3^2$

Thus $AB^2=25$ and $AB=5$ .
The squared distance to $C$ is
$AC^2=5^2+(-2)^2$

This is $29$ , so
$AC=\sqrt{29}>5.$
Therefore $B$ is closer.

Answer

$B$ is closer to $A$ .

Key idea

Straight-line distances can be ordered by comparing their squared coordinate gaps.
Problem 1: Mark complete
Problem 2 From diameter to equation

The equation $(x+2)^2+(y-5)^2=R$ describes a circle whose diameter is $6$ units. Find $R$ .
1. Hint 1
  The radius is half the diameter.
2. Hint 2
  The right side of standard form is the square of the radius.
Answer

$R=9$ .

Full solution

The radius is $6/2=3$ units.
Standard form therefore requires
$R=3^2$

Hence $R=9$ .
The circle of radius $3$ has diameter $2\times3=6$ , as required.

Answer

$R=9$ .

Key idea

The right side of standard form is one quarter of the squared diameter.
Problem 2: Mark complete
Problem 3 A center on a line

A circle has radius $3$ units. Its center lies on the line $y=2x-1$ and has x-coordinate $-2$ . Write the circle equation in standard form.
1. Hint 1
  First determine the missing center coordinate from the line.
2. Hint 2
  Subtract each center coordinate inside its square and square the radius.
Answer

$(x+2)^2+(y+5)^2=9$ .

Full solution

The center height is
$y=2(-2)-1$

This is $-5$ , so the center is $(-2,-5)$ .
Substituting this center and radius $3$ gives
$(x+2)^2+(y+5)^2=9$

Answer

$(x+2)^2+(y+5)^2=9$ .

Key idea

A condition locating the center can be combined with the radius to write a circle equation.
Problem 3: Mark complete
Problem 4 Two boundary points

The figure shows $A=(-1,2)$ and $B=(5,4)$ . A circle through both points has its center on the x-axis. Find its center and write its equation in standard form.

The two points $A$ and $B$ the circle passes through.

Text description of this figure

A coordinate grid. The horizontal x-axis runs from negative three to nine and the vertical y-axis runs from negative five to five, with tick marks, number labels and gridlines at every whole number, equal unit lengths on both axes, and the origin labeled 0. Two points are plotted, each marked with a filled dot and labeled with its letter and its coordinates: A at negative one, two, and B at five, four. Nothing else appears on the grid: no circle, no center, no radius, no midpoint and no construction lines.
1. Hint 1
  Write the center as $(h,0)$ and use equal squared distances to the two boundary points.
2. Hint 2
  Subtract the two distance expressions to find $h$ , then recover the radius squared.
Answer

Center $(3,0)$ ; $(x-3)^2+y^2=20$ .

Full solution

Both points are the same distance from $(h,0)$ , so
$(h+1)^2+4=(h-5)^2+16$

Expand both squares and cancel $h^2$ .
$2h+5=-10h+41$

This gives $h=3$ .
The squared distance to $A$ is $(-1-3)^2+2^2=20$ , so the equation is
$(x-3)^2+y^2=20$

For $B$ , the squared distance is $2^2+4^2=20$ too, confirming that both points lie on the circle.

Answer

Center $(3,0)$ ; $(x-3)^2+y^2=20$ .

Key idea

A center constrained to a line can be found by equating distances to two points on the circle.
Problem 4: Mark complete
Problem 5 Two circular outlines

Two outlines have equations $x^2+y^2+6x-4y+9=0$ and $(x+3)^2+(y-2)^2=9$ . On the blank grid, sketch both circles. State each circle's center and radius, and the difference between the radii.

A blank grid for the two sketches.

Text description of this figure

A blank coordinate grid to sketch on. The horizontal x-axis runs from negative eight to two and the vertical y-axis runs from negative three to seven, with tick marks, number labels and gridlines at every whole number, equal unit lengths on both axes, and the origin labeled 0. Nothing is plotted on the grid: no curves, no points, no center and no labels other than the axis numbers.
1. Hint 1
  Put the first equation in center-radius form by completing both squares.
2. Hint 2
  Compare the two radii, then use four radius steps around each center to sketch that circle.
Answer

Both centers are $(-3,2)$ ; the first circle has radius $2$ and the second radius $3$ ; difference $1$ ; the smaller sketch passes through $(-5,2)$ , $(-1,2)$ , $(-3,0)$ , $(-3,4)$ and the larger through $(-6,2)$ , $(0,2)$ , $(-3,-1)$ , $(-3,5)$ .

Full solution

Group the variable terms in the first equation and move the constant across.
$(x^2+6x)+(y^2-4y)=-9$

Add $9$ and $4$ to both sides to complete the squares, so the right side becomes $-9+9+4$ .
$(x+3)^2+(y-2)^2=4$

The first right side is $4$ , so the first radius is $\sqrt{4}=2$ , and the second right side is $9$ , so the second radius is
$\sqrt{9}=3.$
Both centers are $(-3,2)$ , and their radii differ by $1$ .

For the smaller sketch, use the points $(-5,2)$ , $(-1,2)$ , $(-3,0)$ , and $(-3,4)$ .
For the larger, use $(-6,2)$ , $(0,2)$ , $(-3,-1)$ , and $(-3,5)$ .
Draw round curves through each group.

Answer

Both centers are $(-3,2)$ ; the first circle has radius $2$ and the second radius $3$ ; difference $1$ ; the smaller sketch passes through $(-5,2)$ , $(-1,2)$ , $(-3,0)$ , $(-3,4)$ and the larger through $(-6,2)$ , $(0,2)$ , $(-3,-1)$ , $(-3,5)$ .

Key idea

Completing both squares can reveal that differently written circle equations share a center.
Problem 5: Mark complete
Problem 6 A changed circular sign

A circular sign has boundary $(x-1)^2+(y+2)^2=9$ . A new sign has twice its radius, and its center is $4$ units left and $1$ unit up from the old center. Write the new boundary equation and give its center and radius.
1. Hint 1
  Read the original center and take the square root to find the original radius.
2. Hint 2
  Change the two center coordinates separately, then square the new radius in the equation.
Answer

$(x+3)^2+(y+1)^2=36$ ; center $(-3,-1)$ ; radius $6$ units.

Full solution

The original center is $(1,-2)$ and radius is $3$ .
Moving left four and up one puts the new center at $(-3,-1)$ .
Doubling the radius gives $6$ .

The new standard equation is
$(x+3)^2+(y+1)^2=36$

The center change is $(-4,1)$ and the radius ratio is $6/3=2$ , checking the requested changes.

Answer

$(x+3)^2+(y+1)^2=36$ ; center $(-3,-1)$ ; radius $6$ units.

Key idea

Change a circle’s center and radius before translating them back into its equation.
Problem 6: Mark complete
Problem 7 A changing constant

For each of $k=8$ , $k=10$ , and $k=12$ , describe the real graph of $2x^2+2y^2-8x+4y+k=0$ . For each value, say whether the graph is a circle, a single point, or empty, and give the center and radius, or the point's coordinates, wherever there is one.
1. Hint 1
  Divide by the common leading coefficient before completing both squares.
2. Hint 2
  Inspect whether the resulting right side is positive, zero, or negative for each value.
Answer

$k=8$ : circle, center $(2,-1)$ , radius $1$ . $k=10$ : point $(2,-1)$ . $k=12$ : empty (no real points).

Full solution

Divide by $2$ , then complete the squares in $x$ and $y$ .
$(x-2)^2+(y+1)^2=5-\frac k2$

At $k=8$ , the right side is $1$ , giving a circle centered at $(2,-1)$ with radius $1$ .
At $k=10$ , it is $0$ , forcing both squares to zero and giving the point $(2,-1)$ .
At $k=12$ , it is $-1$ , impossible for a sum of real squares.

Answer

$k=8$ : circle, center $(2,-1)$ , radius $1$ . $k=10$ : point $(2,-1)$ . $k=12$ : empty (no real points).

Key idea

The right side after completing both squares determines whether the graph is a circle, a point, or empty.
Problem 7: Mark complete
Problem 8 A proposed center

A student says that whenever a circle passes through two different points, their midpoint is its center. Is the claim correct? Give a circle and two points that justify your conclusion.
1. Hint 1
  The midpoint is the center when the two points are endpoints of a diameter.
2. Hint 2
  Choose two points on a circle that are not opposite each other.
Answer

False; $x^2+y^2=9$ passes through $(3,0)$ and $(0,3)$ , whose midpoint $(3/2,3/2)$ is not its center.

Full solution

The circle $x^2+y^2=9$ has center $(0,0)$ and radius $3$ .

The points $(3,0)$ and $(0,3)$ both satisfy its equation.
Their midpoint is $(3/2,3/2)$ , which differs from $(0,0)$ .
They are not the ends of a diameter, so their midpoint need not be the center.

Answer

False; $x^2+y^2=9$ passes through $(3,0)$ and $(0,3)$ , whose midpoint $(3/2,3/2)$ is not its center.

Key idea

The midpoint rule locates a circle’s center when the given segment is a diameter.
Problem 8: Mark complete
Problem 9 Inside, on, or outside

A circle has equation $(x+2)^2+(y-3)^2=29$ . Without taking any square root, decide whether each of $(0,8)$ , $(1,1)$ , and $(3,7)$ is inside, on, or outside the circle. Then explain why comparing squared distances gives the same verdict as comparing the distances themselves.
1. Hint 1
  A point's place depends only on how its distance from the center compares with the radius.
2. Hint 2
  Work out each point's squared distance from the center and compare it with the right side.
Answer

$(0,8)$ lies on the circle, $(1,1)$ lies inside it, and $(3,7)$ lies outside it; squaring preserves the order of nonnegative numbers, so the squared comparison gives the same verdict as comparing the distances.

Full solution

The center is $(-2,3)$ and the right side is the radius squared, $29$ .
A point is inside, on, or outside according to whether its squared distance from the center is less than, equal to, or greater than $29$ .
Squaring keeps the order of two nonnegative numbers, and a distance and a radius are both nonnegative, so this squared comparison gives the same verdict as comparing the distance itself with the radius.

For $(0,8)$ the squared distance from the center is
$(0+2)^2+(8-3)^2=29$

That matches the radius squared, so $(0,8)$ lies on the circle.
For $(1,1)$ the squared distance is
$(1+2)^2+(1-3)^2=13$

Since $13<29$ , the point $(1,1)$ lies inside.
For $(3,7)$ the squared distance is
$(3+2)^2+(7-3)^2=41$

Since $41>29$ , the point $(3,7)$ lies outside.

Answer

$(0,8)$ lies on the circle, $(1,1)$ lies inside it, and $(3,7)$ lies outside it; squaring preserves the order of nonnegative numbers, so the squared comparison gives the same verdict as comparing the distances.

Key idea

Comparing a point's squared distance with the radius squared decides inside, on, or outside with no square root.
Problem 9: Mark complete
Problem 10 An altered right side

A circle equation in standard form has right side $16$ . A student adds $9$ to that right side and says the radius increases by $3$ . Is the claim correct? State the actual increase.
1. Hint 1
  The radius is the square root of the entire right side.
2. Hint 2
  Find the two radii separately before subtracting them.
Answer

False; the radius increases by $1$ unit.

Full solution

The original radius is
$\sqrt{16}=4.$
The altered right side is $25$ , with radius
$\sqrt{25}=5.$
The increase is
$5-4=1$

Adding $9$ to the squared radius does not add $\sqrt9$ to the radius.

Answer

False; the radius increases by $1$ unit.

Key idea

A change in radius squared must be interpreted through the new square root, not by adding separate roots.
Problem 10: Mark complete

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