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Equations of 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 A radius fixes the locus

A circle has center $C$ at $(-5,-2)$ and passes through $R$ at $(-3,4)$ . Write its equation and check R in that equation.
1. Hint 1
  Use squared distance from the center to the given point.
2. Hint 2
  Substitute the given boundary point into the equation to check it.
Answer

$(x+5)^2+(y+2)^2=40$ ; R satisfies the equation.

Full solution

The coordinate gaps from C to R are 2 and 6.
The squared radius is
$r^2=2^2+6^2.$
This is 40.

The required equation is
$(x+5)^2+(y+2)^2=40.$
Its left side is the squared distance to C.

Substitute R to check the boundary condition.
$(-3+5)^2+(4+2)^2=40.$
It gives the required squared radius, so R lies on the circle.

The solution triangle uses the horizontal and vertical coordinate gaps.

Answer

$(x+5)^2+(y+2)^2=40$ ; R satisfies the equation.

Key idea

A circle equation imposes a distance condition on every solution point.
Problem 2 Read the geometry and sketch it

Find the center and radius of the graph below. Sketch it and list four circle points horizontally or vertically aligned with its center.
$x^2+y^2+10x-8y+37=0.$
1. Hint 1
  Rewrite the equation so that each variable appears in a completed square.
2. Hint 2
  Step one radius in each coordinate direction from the center.
Answer

Center $(-5,4)$ ; radius 2; four points $(-7,4)$ , $(-3,4)$ , $(-5,2)$ , $(-5,6)$ .

Full solution

Move 37 to the other side and add 25 and 16.
$(x+5)^2+(y-4)^2=4.$
The center is $(-5,4)$ and radius 2.

Stepping left, right, down, and up gives $(-7,4)$ , $(-3,4)$ , $(-5,2)$ , and $(-5,6)$ .
Draw a round curve through these four points on equal-scale axes.

For each listed point, one coordinate gap is zero and the other has magnitude 2.
Its squared distance is therefore 4, verifying the sketch.

The four solution points are each one radius from C.

Answer

Center $(-5,4)$ ; radius 2; four points $(-7,4)$ , $(-3,4)$ , $(-5,2)$ , $(-5,6)$ .

Key idea

A completed-square equation supplies both the center and the scale of a graph.
Problem 3 Which point can be a contact?

A circle has equation
$(x-1)^2+(y+6)^2=17.$
Classify each point as inside, on, or outside. Which point could be the contact point of a tangent to this circle?

A is $(2,-4)$ .

B is $(5,-5)$ .

D is $(-3,-4)$ .
1. Hint 1
  Compare squared distances from the center with the squared radius.
2. Hint 2
  A tangent contact must lie on the circle itself.
Answer

A is inside; B is on the circle; D is outside. Only B can be a tangent contact.

Full solution

For A, the squared center distance is
$1^2+2^2=5.$
Since 5 is less than 17, A is inside.

For B, the squared center distance is
$4^2+1^2=17.$
Thus B is on the circle.

For D, the squared center distance is
$(-4)^2+2^2=20.$
Since 20 exceeds 17, D is outside.

A contact point belongs to the circle, so only B qualifies.
The line perpendicular to the radius at B gives such a tangent.

The solution places A inside, B on the boundary, and D outside.

Answer

A is inside; B is on the circle; D is outside. Only B can be a tangent contact.

Key idea

A contact point passes the circle equation before any line direction is considered.
Problem 4 A vertical chord

Find all common points of the circle and line below. Then find the length and midpoint of the chord between the intersections.
$\begin{aligned}(x+2)^2+(y-1)^2&=18,\\x&=1.\end{aligned}$

The circle and vertical line are given; intersection coordinates are not marked.

Text description of this figure

A vertical line cuts the circle. Both coordinate axes have the same unit scale. C is at (-2, 1). Solid blue circle boundaries and dashed red lines have different line styles.
1. Hint 1
  A vertical line fixes the first coordinate.
2. Hint 2
  The remaining squared equation gives two heights; their difference gives the chord length.
Answer

Intersections $(1,-2)$ and $(1,4)$ ; chord length 6 units; midpoint $(1,1)$ .

Full solution

Substituting the line condition gives
$\begin{aligned}9+(y-1)^2&=18,\\(y-1)^2&=9.\end{aligned}$

The possible heights are negative 2 and 4.
Both points have first coordinate 1, and each makes the circle left side 18.

The chord is vertical, so its length is
$4-(-2)=6.$
Its midpoint is $(1,1)$ , obtained by averaging the two heights.

Answer

Intersections $(1,-2)$ and $(1,4)$ ; chord length 6 units; midpoint $(1,1)$ .

Key idea

A vertical line-circle intersection is solved by substituting x first.
Problem 5 A path through a watered region

On a level plan, each coordinate unit represents one meter. An ideal sprinkler waters every point at distance at most 5 meters from $(3,2)$ . A straight path follows the line below. Find the two boundary crossings and the length of the part of the path inside or on the watered region.
$y=6.$
1. Hint 1
  The boundary is a circle; points on the path must also satisfy its line equation.
2. Hint 2
  Use the two boundary crossings as the endpoints of the watered path segment.
Answer

Boundary crossings $(0,6)$ and $(6,6)$ ; watered path length 6 meters.

Full solution

The ideal circular boundary is
$(x-3)^2+(y-2)^2=25.$
Substitute the path height to obtain
$\begin{aligned}(x-3)^2+16&=25,\\(x-3)^2&=9.\end{aligned}$

The first coordinates are 0 and 6, giving crossings $(0,6)$ and $(6,6)$ .
Along the path, the watered condition is
$(x-3)^2\le9.$
This holds from first coordinate 0 through 6.

The horizontal length is
$6-0=6.$
The scale is one meter per unit, so the requested length is 6 meters.

The solid segment between the solution crossings is the watered part of the dashed path.

Answer

Boundary crossings $(0,6)$ and $(6,6)$ ; watered path length 6 meters.

Key idea

A line condition locates the part of a modeled circular region reached by a path.
Problem 6 One family and one vertical line

For each value of k below, describe the real graph as a positive-radius circle, a single point, or empty. Find all of its common points with $x=6$ . If there is only one common point, explain whether this is tangency to a positive-radius circle.
$x^2+y^2-12x+2y+k=0.$
Case A: $k=21$ .

Case B: $k=37$ .

Case C: $k=40$ .
1. Hint 1
  Complete the two squares before applying the line condition.
2. Hint 2
  A single graph point may lie on a line without forming a positive-radius circle.
Answer

A: circle; $(6,-5)$ and $(6,3)$ . B: single point $(6,-1)$ , not positive-radius circle tangency. C: empty; no common points.

Full solution

Completing the squares gives
$(x-6)^2+(y+1)^2=37-k.$
At first coordinate 6, the first square is zero.

Case A: the right side is 16.
The graph is a circle with center $(6,-1)$ and radius 4.
The two common points are $(6,-5)$ and $(6,3)$ , each 4 units vertically from the center.

Case B: the right side is zero.
Both squares vanish only at $(6,-1)$ , and that point lies on the line.
The graph is a single point with no nonzero contact radius, so this is not tangency to a positive-radius circle.

Case C: the right side is negative 3.
A sum of real squares cannot equal it, so the graph is empty and there are no common points.

Answer

A: circle; $(6,-5)$ and $(6,3)$ . B: single point $(6,-1)$ , not positive-radius circle tangency. C: empty; no common points.

Key idea

One common point implies circle tangency only when the graph is a positive-radius circle.
Problem 7 Check three tangency conditions

Decide whether the line below is tangent to the circle at $T=(1,0)$ . Check that T is on both objects and check the contact-radius direction.
$\begin{aligned}(x+3)^2+(y+2)^2&=20,\\2x+y&=2.\end{aligned}$
1. Hint 1
  A tangent verification needs incidence as well as perpendicularity.
2. Hint 2
  Read the center and compare the slope from it to T with the line slope.
Answer

Yes. T lies on both; the contact-radius slope is $1/2$ and the line slope is $-2$ , so they are perpendicular.

Full solution

At T, the circle left side is
$4^2+2^2=20.$
Thus T is on the circle.

The line left side at T is
$2(1)+0=2.$
Thus T is also on the line.

The center is $C=(-3,-2)$ .
The radius slope is
$m_{CT}=\frac24.$
It is $1/2$ , while the line slope is $-2$ .

Both slopes are defined and nonzero.
Their product is $-1$ , so the line is perpendicular to the radius at its circle endpoint T.
The tangent-radius converse proves tangency.

The right-angle mark is shown only after the direction check.

Answer

Yes. T lies on both; the contact-radius slope is $1/2$ and the line slope is $-2$ , so they are perpendicular.

Key idea

Tangency verification combines a shared point with a perpendicular contact radius.
Problem 8 Compare a line and a circle

Find the perpendicular foot H from the center of the circle to the line below. Use its distance to decide how many common points the line and circle have. Confirm the count by substitution.
$\begin{aligned}x^2+y^2&=10,\\y&=x+6.\end{aligned}$
1. Hint 1
  The center is the origin; its perpendicular to the line has slope negative 1.
2. Hint 2
  Find H from the two line equations, then compare its squared center distance with 10.
Answer

$H=(-3,3)$ ; no common points.

Full solution

The center is $O=(0,0)$ .
The perpendicular through it is
$y=-x.$
Its intersection with the given line satisfies
$-x=x+6.$
Thus $H=(-3,3)$ .

The squared perpendicular distance is
$OH^2=18.$
This exceeds the squared radius 10.
H is the nearest line point to the center, so the entire line is outside.

Substitution confirms the count:
$\begin{aligned}x^2+(x+6)^2&=10,\\2(x+3)^2+18&=10,\\(x+3)^2&=-4.\end{aligned}$
There is no real root, so there are no common points.

The perpendicular foot H lies outside the circle.

Answer

$H=(-3,3)$ ; no common points.

Key idea

A perpendicular foot connects the algebraic intersection test to center distance.
Problem 9 Recover the center from a contact

A circle is tangent to the line below at $T=(3,-1)$ . Its center has first coordinate negative 1. Find its center and equation. Verify the recovered radius and contact direction.
$x-2y=5.$
1. Hint 1
  The center lies on the perpendicular to the tangent through T.
2. Hint 2
  Use the specified first coordinate to find where that perpendicular locates the center.
Answer

Center $(-1,7)$ ; $(x+1)^2+(y-7)^2=80$ .

Full solution

The tangent slope is $1/2$ , so the radius through T has slope $-2$ .
Its line is
$y+1=-2(x-3).$

At first coordinate negative 1, the height is 7.
Thus the center is $C=(-1,7)$ .

The squared radius is
$r^2=4^2+(-8)^2.$
It is 80, so the equation is
$(x+1)^2+(y-7)^2=80.$
The positive radius is $\sqrt{80}$ .

T satisfies the original line because $3-2(-1)=5$ .
It satisfies the recovered circle because its squared center distance is 80.
The radius slope is $-2$ , perpendicular to the tangent slope $1/2$ , checking all the contact conditions.

The vertical dashed line fixes the center first coordinate. The slanted dashed tangent is perpendicular to CT.

Answer

Center $(-1,7)$ ; $(x+1)^2+(y-7)^2=80$ .

Key idea

A known tangent determines the line on which its center must lie.
Problem 10 Combine the circle tests

Use the graph below for all parts. Recover its center and radius. Find every intersection with each of the three lines. For any tangent, verify the perpendicular contact radius. Finally classify $P=(-5,3)$ as inside, on, or outside.
$x^2+y^2+4x-10y+13=0.$
Case A: $x=-7$ .

Case B: $x=-6$ .

Case C: $x=-4$ .
1. Hint 1
  Recover the center-radius equation before testing the lines.
2. Hint 2
  Substitute each first coordinate separately and retain every real height.
3. Hint 3
  A single intersection needs its radius direction checked; compare P by squared distance.
Answer

Center $(-2,5)$ ; radius 4. A: none. B: $(-6,5)$ , tangent with horizontal radius. C: $(-4,5-2\sqrt3)$ and $(-4,5+2\sqrt3)$ . P is inside.

Full solution

Completing the squares gives
$(x+2)^2+(y-5)^2=16.$
The center is $(-2,5)$ and radius 4.

Case A: substitution gives
$(y-5)^2=-9.$
No real height exists, so there are no intersections.

Case B: substitution gives
$(y-5)^2=0.$
The only intersection is $(-6,5)$ .
The radius from $(-2,5)$ to that point is horizontal and the line is vertical, proving tangency.

Case C: substitution gives
$(y-5)^2=12.$
Thus the two heights are $5-2\sqrt3$ and $5+2\sqrt3$ .
Both points satisfy the line and make the circle left side $4+12$ , which is 16.

For P, the squared center distance is
$(-5+2)^2+(3-5)^2=13.$
Since 13 is less than 16, P is inside.

From left to right, dashed lines show Cases A, B, and C. T and the horizontal radius verify Case B.

Answer

Center $(-2,5)$ ; radius 4. A: none. B: $(-6,5)$ , tangent with horizontal radius. C: $(-4,5-2\sqrt3)$ and $(-4,5+2\sqrt3)$ . P is inside.

Key idea

One recovered center-radius equation supports intersection, tangency, and point-position checks.

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