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Circles: 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 The constant entry

The circle $(x+7)^2+(y-4)^2=9$ is written as $x^2+y^2+Dx+Ey+F=0$ . Find $F$ .
1. Hint 1
  Expansion collects the center coordinates and radius into the constant term.
2. Hint 2
  Expand the two squares, then move the right side to the left.
Answer

$F=56$ .

Full solution

The constants from the squares are $49$ and $16$ .
Moving $9$ to the left gives
$F=49+16-9$ $F=56.$
The expanded equation is
$x^2+y^2+14x-8y+56=0.$

Answer

$F=56$ .

Key idea

In general circle form, the constant combines the squared center coordinates with the negative squared radius.
Problem 1: Mark complete
Problem 2 A coefficient record

A circle is recorded as $x^2+y^2+Dx+Ey+F=0$ , with $D=14$ , $E=-6$ , and $F=33$ . Find its center.
1. Hint 1
  The linear terms come from expanding squares centered at the circle's coordinates.
2. Hint 2
  The coefficient of x is $-2h$ , and the coefficient of y is $-2k$ .
Answer

Center $(-7,3)$ .

Full solution

The center coordinates satisfy $-2h=14$ and $-2k=-6$ , so
$h=-7.$
Also
$k=3.$
Completing both squares gives $(x+7)^2+(y-3)^2=25$ , confirming this center and a positive radius.

Answer

Center $(-7,3)$ .

Key idea

The center coordinates are negative one-half of the linear coefficients in normalized general form.
Problem 2: Mark complete
Problem 3 The plotted boundary

The figure shows a circle. Write its equation in center-radius form.

The circle.

Text description of this figure

A grid with the x-axis from -5 to 2 and the y-axis from -3 to 5 on equal scales, a grid line and a label at every integer. A single circle is drawn: its leftmost point is at x equals -4 and its rightmost at x equals 0, both at height 1, and its lowest point is at y equals -1 and its highest at y equals 3, both above x equals -2. No center, radius or point is marked.
1. Hint 1
  Opposite extremes of a circle are equally far from its center.
2. Hint 2
  Read the left and right extremes to find the center's horizontal coordinate and the radius; do the same vertically.
Answer

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

Full solution

The horizontal extremes are $-4$ and $0$ , so their midpoint is $-2$ and the radius is $2$ .
The vertical extremes are $-1$ and $3$ , with midpoint $1$ .
Squaring the distance to $(-2,1)$ gives
$(x+2)^2+(y-1)^2=4.$
Each extreme is $2$ units from the center.

Answer

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

Key idea

A circle's opposite extremes determine its center and radius.
Problem 3: Mark complete
Problem 4 Two possible centers

A circle has radius $\sqrt5$ and passes through $(-8,2)$ . Its center lies on the $x$ -axis. Find all possible centers and their circle equations in center-radius form.
1. Hint 1
  Write a center on the x-axis as $(h,0)$ , and use the given radius as the distance to the known point.
2. Hint 2
  After squaring, isolate $(-8-h)^2$ and retain both signs.
Answer

Centers $(-9,0)$ and $(-7,0)$ ; equations $(x+9)^2+y^2=5$ and $(x+7)^2+y^2=5$ .

Full solution

The squared distance to $(-8,2)$ must be $5$ , so
$(-8-h)^2+4=5.$
Hence
$(h+8)^2=1.$
The two possibilities are $h=-9$ and $h=-7$ , giving the two equations in the answer.
Substituting $(-8,2)$ gives $5$ in each.

Answer

Centers $(-9,0)$ and $(-7,0)$ ; equations $(x+9)^2+y^2=5$ and $(x+7)^2+y^2=5$ .

Key idea

Locating a circle center on a given line may leave two valid centers.
Problem 4: Mark complete
Problem 5 The two rings

The figure shows two circles. Classify their contact as no shared points, internal tangency, external tangency, or two intersections. Justify your choice from their center distance and radii.

Two circles with centers A and B.

Text description of this figure

A grid with the x-axis from -6 to 5 and the y-axis from -5 to 5 on equal scales, a grid line and a label at every integer. A large solid circle has its center marked A on the x-axis at x equals -1; it crosses the x-axis at x equals -5 and x equals 3 and reaches up to y equals 4 and down to y equals -4. A small dashed circle has its center marked B on the x-axis at x equals 2; it crosses the x-axis at x equals 1 and x equals 3. No point where they meet and no radius is marked.
1. Hint 1
  Compare the center distance with both the sum and the positive difference of the radii.
2. Hint 2
  Read each radius from the grid; equality with the difference indicates contact from inside.
Answer

Internal tangency; one shared point.

Full solution

The centers are $(-1,0)$ and $(2,0)$ , so their distance is $d=3$ .
The radii are $4$ and $1$ , giving
$|4-1|=3.$
The center distance equals the radius difference, so the small circle is internally tangent to the large one.
They share exactly one point.

Answer

Internal tangency; one shared point.

Key idea

Distinct circles touch internally when their center distance equals the positive difference of their radii.
Problem 5: Mark complete
Problem 6 A changing coefficient

For real $t$ , consider $x^2+y^2+2tx+16y+68=0$ . Determine which values of $t$ give a circle, a single point, or no real points. Give the point coordinates in the single-point cases.
1. Hint 1
  Complete the squares while treating t as a fixed real coefficient.
2. Hint 2
  Classify the sign of the constant on the right after completing both squares.
Answer

Circle: $t<-2$ or $t>2$ . Single point: $t=2$ gives $(-2,-8)$ ; $t=-2$ gives $(2,-8)$ . No real points: $-2<t<2$ .

Full solution

Completing both squares gives
$(x+t)^2+(y+8)^2=t^2-4.$
The right side is positive for $|t|>2$ , zero for $t=2$ or $t=-2$ , and negative for $|t|<2$ .
A zero sum of two squares forces $(x,y)=(-t,-8)$ , giving the stated two points.

Answer

Circle: $t<-2$ or $t>2$ . Single point: $t=2$ gives $(-2,-8)$ ; $t=-2$ gives $(2,-8)$ . No real points: $-2<t<2$ .

Key idea

The sign of the completed-square constant decides whether a circle-form equation has a circle, one point, or no real graph.
Problem 6: Mark complete
Problem 7 A moving location

A point has coordinates $P=(t,-5)$ , with real $t$ . For which values of $t$ is P strictly inside the circle $x^2+y^2+8x+12y+50=0$ ?
1. Hint 1
  The normalized general-form expression is negative inside the circle.
2. Hint 2
  Substitute the moving point and solve the resulting quadratic inequality.
Answer

$-5<t<-3$ .

Full solution

At P the left side becomes
$t^2+8t+15.$
Its factorization is
$(t+5)(t+3).$
The product is negative between its roots, so P is inside for $-5<t<-3$ .
Completing the circle gives center $(-4,-6)$ and radius $\sqrt2$ ; the line $y=-5$ is $1$ unit from the center, and its points inside satisfy $(t+4)^2+1<2$ , the same interval.

Answer

$-5<t<-3$ .

Key idea

A point is inside a circle when its squared distance from the center is less than the squared radius.
Problem 7: Mark complete
Problem 8 A negative distance

A student squares $\sqrt{(x-6)^2+(y+4)^2}=-3$ and concludes that the original equation describes a circle. Is the conclusion correct? Describe the original graph and the graph obtained after squaring.
1. Hint 1
  Check the signs of both sides before treating squaring as reversible.
2. Hint 2
  Test a point of the squared equation in the original equation.
Answer

No; the original graph is empty. After squaring, the graph is the circle $(x-6)^2+(y+4)^2=9$ , centered at $(6,-4)$ with radius $3$ .

Full solution

The radical denotes the principal square root, which is never negative, so no point satisfies the original equation, and its graph is empty.

Squaring gives $(x-6)^2+(y+4)^2=9$ , the circle with center $(6,-4)$ and radius $3$ .
Its point $(9,-4)$ makes the left side of the original equal $3$ , not $-3$ , so squaring added every one of the circle's points.

Answer

No; the original graph is empty. After squaring, the graph is the circle $(x-6)^2+(y+4)^2=9$ , centered at $(6,-4)$ with radius $3$ .

Key idea

Squaring both sides keeps an equation's solutions unchanged when both sides are nonnegative.
Problem 8: Mark complete
Problem 9 Two records of a boundary

Do $x^2+y^2+12x+8y+45=0$ and $-3x^2-3y^2-36x-24y-135=0$ describe the same circle? Justify your answer and state how many points they share.
1. Hint 1
  Multiplying an equation by a nonzero constant preserves its solutions.
2. Hint 2
  Compare each coefficient, including the constant term.
Answer

Yes; they share infinitely many points.

Full solution

Every coefficient in the second equation is $-3$ times its counterpart in the first.
Dividing the second equation by the nonzero number $-3$ gives the first exactly.
Completing the first equation gives
$(x+6)^2+(y+4)^2=7.$
Both therefore describe the same circle, with every point shared.

Answer

Yes; they share infinitely many points.

Key idea

Nonzero scalar multiples of a circle equation describe the same boundary.
Problem 9: Mark complete
Problem 10 Three locations on a line

A designer requests a circle through $(-2,3)$ , $(1,3)$ , and $(4,3)$ . Is such a circle possible? Justify your answer by explaining what each possible discriminant sign would imply for its intersections with the horizontal line through these points.
1. Hint 1
  The points share one y-coordinate, which can be substituted into a general center-radius equation.
2. Hint 2
  How many real roots can a quadratic in x have?
Answer

No such circle exists. For the substituted quadratic, $\Delta<0$ gives no intersections, $\Delta=0$ gives one, and $\Delta>0$ gives two.

Full solution

Suppose the circle has center $(h,k)$ and radius $r>0$ .
At height $y=3$ , its equation becomes
$(x-h)^2+(3-k)^2=r^2.$
Set
$B=h^2+(3-k)^2-r^2.$
The intersection coordinates must then satisfy
$x^2-2hx+B=0.$
Its discriminant is
$\Delta=4h^2-4B,$
which simplifies to
$\Delta=4[r^2-(3-k)^2].$

If $\Delta<0$ , the line has no intersection with the circle; if $\Delta=0$ , it has one; if $\Delta>0$ , it has two.
The coefficient of $x^2$ is $1$ , so the equation is a quadratic in every case.
None of the cases permits the three distinct x-coordinates $-2,1,4$ , proving that the requested circle does not exist.

Answer

No such circle exists. For the substituted quadratic, $\Delta<0$ gives no intersections, $\Delta=0$ gives one, and $\Delta>0$ gives two.

Key idea

A circle meets a horizontal line at most twice, as shown by the discriminant of the resulting quadratic.
Problem 10: Mark complete

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