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Tangents and Secants: 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 Three line positions

A circle has radius $9$ units. For each line, classify it and give its number of circle intersections.

Case A: perpendicular center distance $12$ units.

Case B: perpendicular center distance $9$ units.

Case C: perpendicular center distance $5$ units.
1. Hint 1
  Compare each perpendicular distance with the radius.
2. Hint 2
  Distances greater than, equal to, or smaller than the radius give different intersection counts.
Answer

Circle-intersection counts:
$\begin{gathered}\text{A: neither; }0,\\\text{B: tangent; }1,\\\text{C: secant; }2.\end{gathered}$

Full solution

Case A: distance $12$ is greater than radius $9$ , so every point of the line is outside.
It has no intersections.

Case B: distance $9$ equals the radius.
Only the perpendicular foot is on the circle, so the line is tangent with one intersection.

Case C: distance $5$ is less than the radius.
There is a circle point on each side of the foot, so the line is secant with two intersections.

Answer

Circle-intersection counts:
$\begin{gathered}\text{A: neither; }0,\\\text{B: tangent; }1,\\\text{C: secant; }2.\end{gathered}$

Key idea

Compare the perpendicular center distance with the radius before classifying a line.
Problem 2 A tangent segment and a radius

A circle has center O and radius $16$ cm. P is $34$ cm from O, and PT is tangent at T. Find the length of segment PT.

Schematic; the requested length is not shown.

Text description of this figure

A schematic circle centered at O has tangent PT, radius OT of 16 cm, and OP of 34 cm. The right angle at T is given by tangency.
1. Hint 1
  Which two segments meet at the contact point?
2. Hint 2
  OP is the hypotenuse of the right triangle OTP. Subtract the radius square from its square.
Answer

$PT=30\text{ cm}$ .

Full solution

The tangent-radius theorem makes OT perpendicular to PT at T.

Pythagoras gives
$\begin{aligned}PT^2&=34^2-16^2,\\PT^2&=900.\end{aligned}$

The length is the positive square root, so
$PT=30\text{ cm}.$

Answer

$PT=30\text{ cm}$ .

Key idea

The center-to-external-point segment is the hypotenuse, not the radius.
Problem 3 Decide whether a line is tangent

T lies on a circle centered at O with radius $9$ cm. Point X differs from T. The given lengths are TX $12$ cm and OX $15$ cm. Decide whether line TX is tangent at T, and justify the decision.
1. Hint 1
  Identify the longest side of triangle OTX.
2. Hint 2
  Compare its square with the sum of the other two squares, then use the location of T.
Answer

Line TX is tangent to the circle at T.

Full solution

The triangle has side lengths $9$ , $12$ , and $15$ cm, with OX longest.

The square comparison is
$\begin{aligned}9^2+12^2&=225,\\15^2&=225.\end{aligned}$

The Pythagorean converse gives a right angle at T, between OT and TX.
Since T is on the circle, the radius-tangent converse makes line TX tangent there.

The right-angle mark appears only in the solution, after the converse establishes it.

Answer

Line TX is tangent to the circle at T.

Key idea

To prove tangency, establish perpendicularity at a point that is actually on the circle.
Problem 4 Recover two equal contact lengths

PA and PB are tangent segments to one circle from the same strict external point P. Their lengths, in centimeters, are $5z-4$ and $2z+11$ . Find z and both segment lengths.

The algebraic lengths are givens; the drawing is schematic.

Text description of this figure

A schematic circle centered at O has tangent contact segments PA and PB from strict external point P. PA is labeled 5z minus 4 and PB is labeled 2z plus 11.
1. Hint 1
  Check the hypotheses for equal tangent segments.
2. Hint 2
  Set the two expressions equal and isolate z. Substitute into both expressions to check positivity.
Answer

$\begin{gathered}z=5,\\PA=21\text{ cm},\\PB=21\text{ cm}.\end{gathered}$

Full solution

The two contact segments come from one strict external point to one circle, so their lengths are equal.

Solve
$\begin{aligned}5z-4&=2z+11,\\3z&=15,\\z&=5.\end{aligned}$

Substituting into the first gives
$PA=21\text{ cm}.$
Substituting into the second also gives
$PB=21\text{ cm}.$
Both lengths are positive.

Answer

$\begin{gathered}z=5,\\PA=21\text{ cm},\\PB=21\text{ cm}.\end{gathered}$

Key idea

A length equation comes from the geometric hypotheses, then each length must be checked.
Problem 5 A straight brace beside a circular window

A circular window is modeled by a circle of radius $20$ cm. A straight brace touches its frame at T and extends to point P outside it. Segment PT is $21$ cm. Assume the brace is tangent and the frame is a perfect circle. Find the distance from P to the window center O.

The ideal-circle and tangent assumptions supply the right angle.

Text description of this figure

A schematic circular frame centered at O has radius OT of 20 cm and a tangent brace segment PT of 21 cm. OP is unknown.
1. Hint 1
  Draw the radius to the contact point.
2. Hint 2
  The given lengths are perpendicular legs. Their squares add to the square of the center distance.
Answer

$OP=29\text{ cm}$ .

Full solution

Under the stated model, the contact radius is perpendicular to the brace.
Triangle OTP is right at T.

Pythagoras gives
$\begin{aligned}OP^2&=20^2+21^2,\\OP^2&=841.\end{aligned}$

Taking the positive square root gives
$OP=29\text{ cm}.$
It exceeds the radius, as required for an external point.

Answer

$OP=29\text{ cm}$ .

Key idea

State the tangency assumption before using a right-triangle model.
Problem 6 Two contact possibilities

Two circles have radii $14$ cm and $6$ cm. Determine their center distance in each case.

Case A: external contact.

Case B: internal contact.

For each case, state the order of the larger center O, smaller center Q, and contact point T on their common line.
1. Hint 1
  At external contact, the contact point separates the centers.
2. Hint 2
  At internal contact, both centers lie on the same side of the tangent at T. Use the radius difference.
Answer

Center distances and point orders, with either order allowed in reverse:
$\begin{gathered}\text{A: }d=20\text{ cm},\\O,T,Q;\\\text{B: }d=8\text{ cm},\\O,Q,T.\end{gathered}$

Full solution

Case A: OT and TQ are radii on opposite sides of T.
The center distance is
$14+6=20\text{ cm}.$
The order is O, T, Q, or its reversal.

Case B: the two radii to T point the same way from their centers.
Subtract their lengths to obtain
$14-6=8\text{ cm}.$
The order is O, Q, T, or its reversal.

The two point orders explain the sum and difference.

Answer

Center distances and point orders, with either order allowed in reverse:
$\begin{gathered}\text{A: }d=20\text{ cm},\\O,T,Q;\\\text{B: }d=8\text{ cm},\\O,Q,T.\end{gathered}$

Key idea

Circle contact uses a sum or a positive difference according to the point order.
Problem 7 Name the common tangent from its centers

A line l touches two circles at distinct points A and B. Their centers O and Q lie on opposite sides of l. The radii are $5$ cm and $2$ cm.

Classify l as a common internal or external tangent.

Could the center distance be $6$ cm? Explain your decision.
1. Hint 1
  Use sides of the tangent line to choose its name.
2. Hint 2
  The two perpendicular radii point in opposite directions, so add their perpendicular separations.
Answer

Line l is a common internal tangent. A center distance of $6$ cm is impossible.

Full solution

Opposite-side centers define a common internal tangent.

The perpendicular separation between the centers is the radius sum,
$5+2=7\text{ cm}.$

The actual center distance cannot be smaller than its perpendicular part.
It must be at least $7$ cm, so $6$ cm is impossible.
Distinct contacts would in fact require a distance strictly greater than $7$ cm.

Answer

Line l is a common internal tangent. A center distance of $6$ cm is impossible.

Key idea

For internal common tangents, the two perpendicular radius distances add.
Problem 8 Justify equal tangent segments

P is a strict external point of a positive-radius circle centered at O. Its two tangent segments touch the circle at distinct points A and B. Prove that PA and PB have equal lengths using triangle congruence. State the congruence criterion and the facts it needs.
1. Hint 1
  Join O to both contact points and to P.
2. Hint 2
  Find the two right angles, the equal radius legs, and the shared hypotenuse.
Answer

$PA=PB$ . The criterion is HL for triangles OAP and OBP.

Full solution

The contact radii are perpendicular to their tangents, so triangles OAP and OBP are right at A and B, respectively.

Their radius legs have equal lengths: $OA=OB$ .

Their hypotenuse OP is the same segment in both triangles.
HL therefore gives $\triangle OAP\cong\triangle OBP$ .

Corresponding sides of the congruent triangles have equal lengths, so $PA=PB$ .

The two right triangles have equal radius legs and a common hypotenuse.

Answer

$PA=PB$ . The criterion is HL for triangles OAP and OBP.

Key idea

A congruence proof identifies both the right triangles and the matching sides.
Problem 9 Separate containment from coincidence

For each pair, give the circle configuration and the total number of common tangent lines.

Case A: radii $8$ and $3$ units, center distance $2$ units.

Case B: radii $8$ and $3$ units, center distance $0$ units.

Case C: radii $8$ and $8$ units, center distance $0$ units.
1. Hint 1
  Check whether the centers coincide before using the ordinary distinct-center cases.
2. Hint 2
  For distinct centers compare the distance with the radius difference; for the same center compare radii.
Answer

Common-tangent counts:
$\begin{gathered}\text{A: nested; }0,\\\text{B: concentric; }0,\\\text{C: coincident; }\infty.\end{gathered}$
Here infinity means infinitely many distinct common tangent lines.

Full solution

Case A: the radius difference is $5$ units.
Since the positive center distance $2$ is smaller, the smaller circle is strictly nested with no contact.
There are no common tangents.

Case B: the circles are concentric with unequal radii.
One line cannot have two different perpendicular distances from the shared center, so there are no common tangents.

Case C: equal center and equal radius give coincident circles.
Every tangent to one is a tangent to the other, so there are infinitely many.

Answer

Common-tangent counts:
$\begin{gathered}\text{A: nested; }0,\\\text{B: concentric; }0,\\\text{C: coincident; }\infty.\end{gathered}$
Here infinity means infinitely many distinct common tangent lines.

Key idea

An equal-center case needs its own radius comparison.
Problem 10 Identify a one-tangent configuration

Two circles have radii $8$ cm and $3$ cm. Their centers are distinct. There is exactly one distinct common tangent line.

Find the center distance, state the kind of contact, and classify the common tangent.

The smaller circle is then moved directly away from the larger center until the center distance is $11$ cm. State the new kind of contact and the new numbers of external and internal common tangents.
1. Hint 1
  Use the complete common-tangent count table to locate the one-line case.
2. Hint 2
  With unequal radii, compare the center distance with both their difference and their sum.
3. Hint 3
  At the sum, count the contact tangent once.
Answer

Initially:
$\begin{gathered}d=5\text{ cm},\\\text{internal contact},\\1\text{ external tangent}.\end{gathered}$
At 11 cm:
$\begin{gathered}\text{external contact},\\2\text{ external tangents},\\1\text{ internal tangent}.\end{gathered}$

Full solution

Initially, distinct-center circles have exactly one common tangent only at internal contact.
The center distance is the positive radius difference:
$8-3=5\text{ cm}.$

At internal contact both centers lie on the same side of the shared tangent, so that tangent is external.

After the move, the center distance $11$ cm equals the radius sum.
The circles touch externally.

The two common external tangents exist.
The two internal tangents have merged into one line at the contact, so the new counts are two external and one internal.

Answer

Initially:
$\begin{gathered}d=5\text{ cm},\\\text{internal contact},\\1\text{ external tangent}.\end{gathered}$
At 11 cm:
$\begin{gathered}\text{external contact},\\2\text{ external tangents},\\1\text{ internal tangent}.\end{gathered}$

Key idea

A tangent count can determine the two-circle configuration before any diagram is drawn.

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