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Circle Geometry: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Core practice

10 problems from across the chapter. 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 successive maps

Three circles have distinct centers $O,P,Q$ and positive radii $8,14,35$ , respectively. Map $F$ translates $O$ to $P$ and then dilates about $P$ with a positive scale factor. Map $G$ translates $P$ to $Q$ and then dilates about $Q$ with a positive scale factor. Each map sends its entire source circle onto its entire target circle.

Find the dilation factor in $F$ , the dilation factor in $G$ , and the total length scale factor in $G\circ F$ . Explain why $G\circ F$ is one-to-one and onto the entire third circle, using an arbitrary point of that circle.
1. Hint 1
  A translation preserves the distance of each point from the translated center.
2. Hint 2
  Compare target and source radii at each dilation, then multiply the factors for the composition.
3. Hint 3
  Start with an arbitrary point on the third circle and undo the two maps in reverse order.
Answer

$\begin{gathered}k_F=\frac74,\qquad k_G=\frac52\\ k_{G\circ F}=\frac{35}{8}\end{gathered}$
The composition is one-to-one and onto.

Full solution

After the first translation, each point is $8$ units from $P$ .
To make that distance $14$ , use
$k_F=14/8=7/4.$
Likewise,
$k_G=35/14=5/2.$

Translations have length scale factor $1$ .
The composition therefore has total factor
$\frac74\cdot\frac52=\frac{35}{8}.$

Take any point $Y$ on the third circle.
Dilate $Y$ about $Q$ by $2/5$ and undo the translation from $P$ to $Q$ .
This gives a point on the second circle, since its distance from $P$ is $14$ .
Then dilate about $P$ by $4/7$ and undo the translation from $O$ to $P$ .
The resulting point is $8$ units from $O$ , so it belongs to the first circle and maps to $Y$ .
This proves onto for every target point.

Each inverse translation and each positive inverse dilation is uniquely defined.
Thus $Y$ has exactly one preimage.
The composition is one-to-one as well as onto.

Answer

$\begin{gathered}k_F=\frac74,\qquad k_G=\frac52\\ k_{G\circ F}=\frac{35}{8}\end{gathered}$
The composition is one-to-one and onto.

Key idea

Positive radius ratios give invertible whole-circle maps, and composition multiplies their length scale factors.
Problem 2 Positions of two circles

Cases A, B and C concern two circles of radii $16$ and $7$ , with center distance $d\geq0$ . For every tangent count, distinguish external common tangents, whose centers lie on the same side of the line, from internal common tangents, whose centers lie on opposite sides. Count distinct lines.

Case A: $d=9$ . Classify the circle contact and give both common-tangent counts.

Case B: $d=23$ . Classify the circle contact and give both common-tangent counts.

Case C: Find all center distances $d$ for which the two circles have exactly two common tangents in total.

Case D: A separate pair consists of two circles of radius $16$ with the same center. Classify the pair and give both common-tangent counts.
1. Hint 1
  Compare $d$ with the radius difference and the radius sum.
2. Hint 2
  At a contact point, decide which side of the common tangent contains each center. Count a shared contact line only once.
3. Hint 3
  For case C, distinguish the open intersecting range from both contact boundaries. For case D, compare the tangent-line sets of circles with the same center and radius.
Answer

$\begin{gathered}\text{A: internal contact }(1,0)\\ \text{B: external contact }(2,1)\\ \text{C: }9<d<23\\ \text{D: coincident }(\infty,0)\end{gathered}$
Pairs count external, then internal tangents.

Full solution

The critical distances are $16-7=9$ and $16+7=23$ .

Case A has internal contact: the smaller circle is inside the larger and they share one point.
Their sole common tangent has both centers on the same side.
There is one external common tangent and no internal common tangent.

Case B has external contact.
There are two external common tangents, plus the contact tangent with the centers on opposite sides.
Thus the external and internal counts are $2$ and $1$ , respectively.

Case C requires two intersection points.
This occurs exactly when
$9<d<23.$
There are then two external common tangents and no internal ones.
At the two endpoints the totals are $1$ and $3$ , so neither endpoint belongs to the interval.
For $d<9$ there are no common tangents, and for $d>23$ there are four.

Case D consists of the same circle twice.
Every tangent to that circle is common to both, giving infinitely many external common tangents.
The coincident centers cannot lie on opposite sides of any line, so there are no internal common tangents.

Answer

$\begin{gathered}\text{A: internal contact }(1,0)\\ \text{B: external contact }(2,1)\\ \text{C: }9<d<23\\ \text{D: coincident }(\infty,0)\end{gathered}$
Pairs count external, then internal tangents.

Key idea

The radius difference and sum locate the contact boundaries, and common tangents are counted as distinct lines.
Problem 3 A diameter and two rays

A circle has center $O$ and diameter $AB$ . Point $C$ is on the circle and is distinct from $A,B$ , with $m\angle CAB=31^\circ$ . The tangent at $A$ contains points $R,S$ on opposite rays from $A$ . Point $R$ and point $C$ lie on the same side of line $OA$ .

Find $m\angle ACB$ , $m\angle CBA$ , $m\angle RAC$ and $m\angle SAC$ . For each angle involving a tangent ray, name the intercepted arc and give its measure. Justify the angle choices.

Text description of this figure

AB is a diameter. The tangent rays AR and AS are opposite, and R is on the same side of OA as C.
1. Hint 1
  Determine the angle subtended by the diameter before completing triangle ABC.
2. Hint 2
  Use the inscribed angle at B to find the arc AC that excludes B.
3. Hint 3
  Ray AR is the ray toward the side containing C. The two tangent-chord angles form a linear pair.
Answer

$\begin{gathered}\angle ACB=90^\circ,\quad\angle CBA=59^\circ\\ \angle RAC=59^\circ,\quad\angle SAC=121^\circ\\ \widehat{AC}_{\rm minor}=118^\circ\\ \widehat{ABC}_{\rm major}=242^\circ\end{gathered}$

Full solution

Because $AB$ is a diameter, the semicircle theorem gives
$m\angle ACB=90^\circ.$
The remaining triangle angle is
$m\angle CBA=180^\circ-90^\circ-31^\circ=59^\circ.$

Angle $CBA$ intercepts the arc $AC$ excluding $B$ , so that minor arc measures $118^\circ$ .
The other arc, through $B$ , measures
$360^\circ-118^\circ=242^\circ.$

The radius $OA$ is perpendicular to the tangent.
Since $R$ is on the side containing $C$ ,
$m\angle RAC=90^\circ-31^\circ=59^\circ.$
It intercepts the minor arc $AC$ , agreeing with half of $118^\circ$ .

The opposite tangent ray gives the supplementary angle
$m\angle SAC=180^\circ-59^\circ=121^\circ.$
It intercepts the major arc $ABC$ , agreeing with half of $242^\circ$ .

Answer

$\begin{gathered}\angle ACB=90^\circ,\quad\angle CBA=59^\circ\\ \angle RAC=59^\circ,\quad\angle SAC=121^\circ\\ \widehat{AC}_{\rm minor}=118^\circ\\ \widehat{ABC}_{\rm major}=242^\circ\end{gathered}$

Key idea

A tangent line determines two ray angles, and their intercepted arcs are the two different branches between the chord endpoints.
Problem 4 An interior intersection

Chords $AB$ and $CD$ of one circle meet at a point $P$ strictly inside it, in the orders $A,P,B$ and $C,P,D$ . The four endpoints are distinct. Given $PA=11$ , $PB=16$ and $PC=8$ , find $PD$ , the signed power of $P$ and the signed power of $A$ .

Justify the length calculation with a correctly ordered triangle similarity statement and the angle equalities that establish it.

Text description of this figure

Two chords meet at P inside the circle, with the endpoint orders given in the statement.
1. Hint 1
  Compare triangle APC with a triangle using D, P and B. Match the equal angles before writing a proportion.
2. Hint 2
  The angles at P are vertical. Another angle pair intercepts the same arc BC.
3. Hint 3
  An interior product is positive, but interior signed power is negative. Consider the location of A separately from the strictly interior location of P.
Answer

$\begin{gathered}PD=22\\ \operatorname{Pow}(P)=-176\\ \operatorname{Pow}(A)=0\end{gathered}$

Full solution

Draw chords $AC$ and $DB$ .
The angles $APC$ and $DPB$ are vertical and equal.
Since rays $AP$ and $AB$ agree, and rays $DP$ and $DC$ agree, angles $PAC$ and $PDB$ both intercept arc $BC$ and are equal.
Thus
$\triangle APC\sim\triangle DPB.$

The correspondence is $A\leftrightarrow D$ , $P\leftrightarrow P$ , $C\leftrightarrow B$ .
It gives
$\begin{gathered}\frac{PA}{PD}=\frac{PC}{PB}\\ \frac{11}{PD}=\frac8{16}\\ PD=22\end{gathered}$
Equivalently, both chord products are $176$ .

Since $P$ is strictly inside, its signed power is the negative of that product:
$\operatorname{Pow}(P)=-176.$
Point $A$ lies on the circle, so
$\operatorname{Pow}(A)=0.$

Answer

$\begin{gathered}PD=22\\ \operatorname{Pow}(P)=-176\\ \operatorname{Pow}(A)=0\end{gathered}$

Key idea

Match corresponding vertices before forming a product, then distinguish that positive product from signed power.
Problem 5 Four angles in a quadrilateral

A convex quadrilateral $ABCD$ has distinct vertices in boundary order, with no three collinear. Diagonals $AC$ and $BD$ meet at $X$ . Given $m\angle BAD=108^\circ$ , $m\angle BCD=72^\circ$ , $m\angle BAC=35^\circ$ and $m\angle ABD=24^\circ$ , determine whether one circle passes through all four vertices and find $m\angle BXC$ . Justify the conclusion and the angle transfers.

Text description of this figure

The convex quadrilateral and both diagonals are shown; cyclicity is to be determined.
1. Hint 1
  Use the two given whole angles to decide whether the opposite-angle converse applies.
2. Hint 2
  After cyclicity is established, compare angles $ABD$ and $ACD$ , which subtend the same chord.
3. Hint 3
  Find $\angle BCX$ and $\angle CBX$ separately, then use triangle $BXC$ .
Answer

The quadrilateral is cyclic, and $m\angle BXC=59^\circ$ .

Full solution

The opposite whole angles satisfy
$108^\circ+72^\circ=180^\circ.$
The convex boundary order and distinctness hypotheses allow the converse cyclicity test, so all four vertices lie on one circle.

Now angles $ACD$ and $ABD$ intercept the same arc $AD$ , so
$m\angle ACD=24^\circ.$
The diagonal splits the whole angle at $C$ , giving
$m\angle BCA=72^\circ-24^\circ=48^\circ.$

At $A$ ,
$m\angle CAD=108^\circ-35^\circ=73^\circ.$
The same-arc equality gives
$m\angle CBD=73^\circ.$
Since $X$ lies on both diagonals, these are the angles $BCX$ and $CBX$ .

Triangle $BXC$ therefore gives
$\begin{gathered}m\angle BXC=180^\circ-48^\circ-73^\circ\\ =59^\circ\end{gathered}$

Answer

The quadrilateral is cyclic, and $m\angle BXC=59^\circ$ .

Key idea

Establish cyclicity before transferring diagonal angles by equal intercepted arcs.
Problem 6 Angles across one circle

Distinct points $A,B,C,D$ occur in that order on a circle. Tangents at $A$ and $C$ meet at an external point $P$ , with $B$ on the minor arc $AC$ . Diagonals $AC$ and $BD$ meet at $X$ . Given $m\angle APC=82^\circ$ , $m\angle AXB=93^\circ$ and consecutive arc $BC$ of measure $47^\circ$ , find $m\angle BAD$ . Determine all four consecutive arc measures to verify that the information gives positive arcs filling one circle.

Text description of this figure

The circle order is A, B, C, D, with B on the minor contact arc AC.
1. Hint 1
  The two-tangent angle determines the minor contact arc ABC.
2. Hint 2
  The inside angle $AXB$ uses consecutive arcs $AB$ and $CD$ .
3. Hint 3
  Once all consecutive arcs are known, select the arc BD that excludes A.
Answer

$\begin{gathered}\widehat{AB}=51^\circ,\quad\widehat{BC}=47^\circ\\ \widehat{CD}=135^\circ,\quad\widehat{DA}=127^\circ\\ m\angle BAD=91^\circ\end{gathered}$

Full solution

The minor contact arc is
$m\widehat{ABC}=180^\circ-82^\circ=98^\circ.$
Removing arc $BC$ gives
$m\widehat{AB}=98^\circ-47^\circ=51^\circ.$

For the inside angle,
$93^\circ=\frac{m\widehat{AB}+m\widehat{CD}}2.$
Therefore
$m\widehat{CD}=186^\circ-51^\circ=135^\circ.$

The remaining arc is $m\widehat{DA}=360^\circ-51^\circ-47^\circ-135^\circ=127^\circ$ .
All four measures are positive and their sum is $360^\circ$ , while $AB+BC=98^\circ<180^\circ$ verifies the specified minor branch.

Angle $BAD$ intercepts arc $BCD$ , which excludes $A$ .
Consequently
$m\angle BAD=\frac{47^\circ+135^\circ}{2}=91^\circ.$

Answer

$\begin{gathered}\widehat{AB}=51^\circ,\quad\widehat{BC}=47^\circ\\ \widehat{CD}=135^\circ,\quad\widehat{DA}=127^\circ\\ m\angle BAD=91^\circ\end{gathered}$

Key idea

Linked tangent and interior angles determine consecutive arcs before an inscribed angle selects its vertex-excluded branch.
Problem 7 A concentric circle

A circle with center $O$ and radius $\sqrt{274}$ has distinct chords $AB$ and $CD$ of lengths $30$ and $24$ , respectively. A second circle with the same center and positive radius is tangent to line $AB$ . Find its radius and the number of intersections of line $CD$ with this second circle. Justify both conclusions.

Also determine whether a line containing a diameter of the original circle could be tangent to any positive-radius circle with center $O$ . Explain the boundary case.

Text description of this figure

Only the original circle with center O and chords AB and CD is shown; the new circle is to be determined.
1. Hint 1
  A perpendicular from the center bisects each chord. Find the distance from O to each chord line.
2. Hint 2
  For the new circle, compare each line's center distance with its radius.
3. Hint 3
  A diameter line passes through the common center, so its center distance is zero.
Answer

The second radius is $7$ . Line $CD$ has $0$ intersections with it. A diameter line is a secant of every positive-radius circle centered at $O$ , so it cannot be tangent.

Full solution

Let $H$ and $K$ be the perpendicular feet from $O$ to lines $AB$ and $CD$ .
Chord bisection gives half-lengths $15$ and $12$ .
Thus
$\begin{gathered}OH=\sqrt{274-15^2}=7\\ OK=\sqrt{274-12^2}=\sqrt{130}\end{gathered}$

A tangent line has center distance equal to the radius.
The second circle therefore has radius $7$ .
Conversely, its perpendicular foot $H$ lies on the circle and $OH$ is perpendicular to $AB$ , verifying tangency.

For line $CD$ , the center distance $\sqrt{130}$ is greater than $7$ .
The line stays outside the second circle and has no intersections with it.

A diameter line has center distance $0$ .
Every positive-radius circle centered at $O$ has two intersections with a line through $O$ .
Thus the diameter line is a secant, never a tangent.
A zero-radius object would not be a circle in this problem.

Answer

The second radius is $7$ . Line $CD$ has $0$ intersections with it. A diameter line is a secant of every positive-radius circle centered at $O$ , so it cannot be tangent.

Key idea

Center distance determines whether a line misses, touches or crosses a circle, including a line through its center.
Problem 8 Diagonals and a length

A convex quadrilateral $ABCD$ has four distinct vertices in boundary order, with no three collinear. Its opposite whole angles satisfy $m\angle BAD=m\angle BCD=90^\circ$ . Diagonals $AC$ and $BD$ meet at $X$ , with $AX=7$ , $XC=12$ and $BX=6$ . Determine whether one circle passes through all four vertices, justify that conclusion, and then find $XD$ and the signed power of $X$ with respect to that circle.

Text description of this figure

The right-angle marks show the given whole angles at A and C.
1. Hint 1
  First use the opposite whole angles and the convexity hypotheses to decide cyclicity.
2. Hint 2
  After the circle is established, AC and BD are chords through its interior point X.
3. Hint 3
  Check the sign of power separately from the positive segment product.
Answer

The quadrilateral is cyclic.
$XD=14,\qquad\operatorname{Pow}(X)=-84.$

Full solution

The opposite whole angles sum to $180^\circ$ .
The converse opposite-angle test applies to the distinct convex ordered vertices, so a circle passes through $A,B,C,D$ .
This conclusion follows from the angles before any segment-product rule is used.

The diagonals of a convex quadrilateral cross inside it, so $X$ is an interior point of the circle.
The endpoint orders are $A,X,C$ and $B,X,D$ .
Therefore
$\begin{gathered}AX\cdot XC=BX\cdot XD\\ 7\cdot12=6\cdot XD\\ XD=14\end{gathered}$

Both products equal $84$ .
Since $X$ is inside, its signed power is
$\operatorname{Pow}(X)=-84.$

Answer

The quadrilateral is cyclic.
$XD=14,\qquad\operatorname{Pow}(X)=-84.$

Key idea

An angle converse can establish the circle needed before an intersecting-chord product is available.
Problem 9 Three consecutive arcs

A tangent from an external point $P$ touches a circle at $T$ . A secant has order $P,A,B$ , with $T,A,B$ distinct and in that order around the circle. Let $n$ be the measure of arc $TA$ excluding $B$ , $b$ the measure of arc $AB$ excluding $T$ , and $f$ the measure of arc $BT$ excluding $A$ . Given $m\angle TPA=65^\circ$ and $f=2b$ , determine $n,b,f$ . Verify positivity and the full-circle sum.

Point $Q$ is on the secant line with order $Q,P,A,B$ . Find $m\angle TPQ$ and explain how the actual ray changes the angle.

Text description of this figure

The secant ray PA and its opposite ray PQ are both shown, together with tangent segment PT.
1. Hint 1
  For the stated tangent and secant rays, the angle is half the far arc minus the near arc.
2. Hint 2
  Use both $n+b+f=360$ and $f=2b$ to reduce the arc equations to one unknown.
3. Hint 3
  Rays PA and PQ are opposite. Check the linear pair only after determining the secant-ray angle.
Answer

$\begin{gathered}n=66^\circ,\quad b=98^\circ,\quad f=196^\circ\\ m\angle TPQ=115^\circ\end{gathered}$

Full solution

The tangent-secant exterior rule uses the stated branches:
$65=\frac{f-n}{2},\qquad f-n=130.$
The far arc need not be minor.

Using $f=2b$ gives $n=2b-130$ .
Substitute in the full-circle equation:
$\begin{gathered}(2b-130)+b+2b=360\\ 5b=490\\ b=98\end{gathered}$
Then $f=196$ and $n=66$ , with all measures in degrees.

The measures are positive, their sum is $66+98+196=360$ , and $(196-66)/2=65$ checks the actual exterior angle.
Thus the major far arc is consistent with the secant configuration.

Rays $PA$ and $PQ$ are opposite, so
$m\angle TPQ=180^\circ-65^\circ=115^\circ.$
Using the same arc difference for this reversed ray without taking the supplement would give the wrong branch.

Answer

$\begin{gathered}n=66^\circ,\quad b=98^\circ,\quad f=196^\circ\\ m\angle TPQ=115^\circ\end{gathered}$

Key idea

Exterior arc equations must agree with the full circle and with the actual rays defining the requested angle.
Problem 10 A chord between contacts

From a point $P$ strictly outside one circle, segments $PT$ and $PU$ are tangent at distinct points $T,U$ , with $m\angle TPU=120^\circ$ . A secant through the same point has order $P,A,B$ , with $PA=5$ and $AB=15$ . Find the measure of the minor arc $TU$ , the chord length $TU$ and the signed power of $P$ . Justify all three results.

Text description of this figure

The tangents, secant and requested contact chord belong to the same circle.
1. Hint 1
  The two-tangent angle determines the minor contact arc.
2. Hint 2
  Use the external and whole secant lengths to find a tangent length, then compare the two finite tangent segments from P.
3. Hint 3
  Drop a perpendicular from P to chord TU in the resulting isosceles triangle. Its two right triangles have familiar angles.
Answer

$\begin{gathered}m\widehat{TU}_{\rm minor}=60^\circ\\ TU=10\sqrt3\\ \operatorname{Pow}(P)=100\end{gathered}$

Full solution

The two-tangent angle is $180^\circ$ minus the minor contact arc.
Hence
$\begin{gathered}m\widehat{TU}_{\rm minor}=180^\circ-120^\circ\\ =60^\circ\end{gathered}$

The whole secant is
$PB=5+15=20.$
The tangent-secant product gives $PT^2=5\cdot20=100$ , so $PT=10$ .
Finite tangents from the same external point are equal, giving $PU=10$ as well.

Triangle $PTU$ is therefore isosceles with vertex angle $120^\circ$ .
Drop the perpendicular $PM$ to base $TU$ .
In an isosceles triangle this altitude also bisects the base and vertex angle, so $m\angle TPM=60^\circ$ , $m\angle PTM=30^\circ$ and $PT=10$ .

In the $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, $PM=5$ and
$TM=5\sqrt3.$
Thus
$TU=2TM=10\sqrt3.$

Since $P$ is outside, its signed power is the positive secant product
$\operatorname{Pow}(P)=5\cdot20=100.$
All length calculations concern this same circle and the given secant order.

Answer

$\begin{gathered}m\widehat{TU}_{\rm minor}=60^\circ\\ TU=10\sqrt3\\ \operatorname{Pow}(P)=100\end{gathered}$

Key idea

A contact angle, a secant product and equal finite tangents can connect arc measure to a chord length.

Multiple choice

Core practice