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Constructing Tangents from an External Point: 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: Advanced (beyond the core course) Advanced. This problem set goes beyond core Geometry. You can skip it.

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Problem 1 of 10

Problem 1 Finish the compass setup

A circle of radius $9$ cm has center $O$ . The outside point $P$ is $26$ cm from $O$ .

Two pairs of arcs drawn with one fixed compass opening from $O$ and $P$ have already met at $U$ and $V$ , as shown.

Describe how to locate the auxiliary circle's center. State the compass opening for that circle and explain why the arc crossings determine its center.

The compass arcs are already in place.

Text description of this figure

Arcs with equal openings from O and P meet at U and V. Segment OP is drawn but its midpoint is unmarked.
1. Hint 1
  Each arc crossing is a fixed compass distance from both endpoints.
2. Hint 2
  Find the line that contains both arc crossings, then use its intersection with the segment.
Answer

Draw line $UV$ and take its crossing $M$ with $OP$ as the center. The compass opening is $13$ cm.

Full solution

The fixed compass opening gives equal distances from $U$ to $O$ and $P$ .
It also gives equal distances from $V$ to $O$ and $P$ .

Both arc crossings lie on the perpendicular bisector of $OP$ .
Draw line $UV$ to locate its crossing $M$ with $OP$ .

Open the compass from $M$ to $O$ .
The opening is half of $26$ cm, or $13$ cm.

Join U and V before setting the compass opening for the dashed circle.

Answer

Draw line $UV$ and take its crossing $M$ with $OP$ as the center. The compass opening is $13$ cm.

Key idea

Two points equally far from the same endpoints locate the perpendicular bisector.
Problem 2 Use the two circles in the drawing

The diagram shows a given circle centered at $O$ and the auxiliary circle with diameter $OP$ . Point $P$ is outside the given circle. Their crossings are $T_1$ and $T_2$ .

Name the two lines to draw through $P$ . Explain separately how the auxiliary circle and the given circle establish that the line through $T_2$ is tangent.

Both circles and their two crossings are drawn.

Text description of this figure

The solid circle is centered at O. The dashed circle is centered at M and passes through O and P. Both T₁ and T₂ are on the circles.
1. Hint 1
  The two ends of the auxiliary diameter are on the rays of the angle with vertex $T_2$ .
2. Hint 2
  Once you have a right angle, identify which of its sides lies along a radius of the given circle.
Answer

Draw line $PT_1$ and line $PT_2$ . The auxiliary circle gives a right angle at $T_2$ , where $OT_2$ is a radius of the given circle.

Full solution

Draw one line through $P$ and $T_1$ and a second line through $P$ and $T_2$ .

For the auxiliary circle, $OP$ is a diameter and $T_2$ is on the circle.
The semicircle theorem makes $\angle OT_2P$ a right angle.

For the given circle, $OT_2$ is a radius because $T_2$ is on that circle.
Line $PT_2$ is perpendicular to the radius at its endpoint, so it is tangent there.

The two completed lines and the right angles they make with the contact radii.

Answer

Draw line $PT_1$ and line $PT_2$ . The auxiliary circle gives a right angle at $T_2$ , where $OT_2$ is a radius of the given circle.

Key idea

One circle supplies the right angle and the other supplies the contact radius.
Problem 3 Find the compass opening and the tangent lengths

The given circle has radius $17$ cm and center $O$ . The outside point $P$ is $33$ cm from $O$ .

Find the auxiliary compass opening and the exact lengths of both tangent segments from $P$ to their contact points $T_1$ and $T_2$ .
1. Hint 1
  The segment joining the given center and $P$ serves as a diameter of one circle and a hypotenuse of two triangles.
2. Hint 2
  Use the midpoint for the compass opening and the Pythagorean theorem for the tangent lengths.
Answer

$\begin{gathered}MO=16.5\text{ cm}\\PT_1=20\sqrt{2}\text{ cm}\\PT_2=20\sqrt{2}\text{ cm}.\end{gathered}$

Full solution

The auxiliary radius is half the diameter $OP$ , so the opening is $16.5$ cm.

Each constructed line is perpendicular to its contact radius.
In either triangle the hypotenuse is $OP$ , and the radius is a leg.

For either contact $T$ , Pythagoras gives
$\begin{aligned}PT^2&=33^2-17^2\\&=800\\PT&=20\sqrt{2}\text{ cm}.\end{aligned}$

The positive root gives the length for each of the two contacts.

Use the same hypotenuse and radius in either triangle.

Answer

$\begin{gathered}MO=16.5\text{ cm}\\PT_1=20\sqrt{2}\text{ cm}\\PT_2=20\sqrt{2}\text{ cm}.\end{gathered}$

Key idea

The tangent triangles have a common hypotenuse and equal contact radii.
Problem 4 Write the equation of a boundary tangent

A circle of radius $5$ units is centered at $O=(2,-1)$ . The point $P=(6,2)$ lies on the circle.

Find an equation of the tangent line through $P$ . Explain why no other tangent passes through $P$ .
1. Hint 1
  If a tangent passes through a point that is already on the circle, that point must be its contact.
2. Hint 2
  Find the slope of $OP$ and take its negative reciprocal. Use the coordinates of $P$ in the line equation.
Answer

The unique tangent is $y-2=-\frac43(x-6)$ , or any equivalent equation.

Full solution

A tangent through $P$ has $P$ as its contact point and must be perpendicular to $OP$ there.

The horizontal change from $O$ to $P$ is $4$ and the vertical change is $3$ .
The radius slope is $\frac34$ .

The perpendicular slope is $-\frac43$ .
Using $P$ gives
$y-2=-\frac43(x-6).$

There is exactly one perpendicular to $OP$ through $P$ , so it gives the unique tangent.

The tangent direction is determined by the radius at P.

Answer

The unique tangent is $y-2=-\frac43(x-6)$ , or any equivalent equation.

Key idea

A boundary point already supplies the contact point.
Problem 5 Compare positions on opposite rays

A circle has center $O$ and radius $23$ units. The point $A$ lies on the circle.

Case A. Start from $A$ itself.

Case B. Start from $B$ , on the ray opposite $OA$ , at a distance of $20$ units from $O$ .

Case C. Start from $C$ , on ray $OA$ and $4$ units beyond $A$ .

Let $n_A$ , $n_B$ , and $n_C$ be the tangent counts in Cases A, B, and C. Find each count and explain how you decided.

The labeled points lie on a common line through the center.

Text description of this figure

A is where the right ray from O meets the circle. C lies farther along that ray. B is on the left ray from O.
1. Hint 1
  For each starting point, compare its distance from $O$ with the radius.
2. Hint 2
  Changing the direction of a ray changes its side of the center. It does not change how far along that ray the circle is.
Answer

$\begin{gathered}n_A=1\\n_B=0\\n_C=2.\end{gathered}$

Full solution

Case A.
Since $A$ lies on the circle, the single tangent is the perpendicular to $OA$ at $A$ .

Case B.
The distance $OB$ is $20$ units, smaller than the radius.
Every line through this interior point crosses the circle twice, so there is no tangent.

Case C.
The distance $OC$ is $27$ units.
The point is outside the circle, so the circle with diameter $OC$ gives two contacts and two tangents.

Answer

$\begin{gathered}n_A=1\\n_B=0\\n_C=2.\end{gathered}$

Key idea

The center distance determines the tangent count on either ray.
Problem 6 Recover the contacts from one surviving line

A diagram shows a positive-radius circle centered at $O$ , an outside point $P$ , and one tangent line through $P$ . Its point of contact has not been marked.

Using a compass and straightedge, describe how to find that contact point and construct the other tangent. Explain why the other line must be different.

Start with the given tangent and the unmarked point where it touches the circle.

Text description of this figure

A solid circle is centered at O. One tangent line through P is drawn, but its contact point is not marked.
1. Hint 1
  The original contact point is the foot of a perpendicular from $O$ .
2. Hint 2
  Use $OP$ as a mirror line after locating the first contact point. To reflect a point, copy its perpendicular distance to the opposite side.
Answer

Drop a perpendicular from $O$ to the given line to find $T_1$ . Reflect $T_1$ across $OP$ to obtain $T_2$ , then draw line $PT_2$ .

Full solution

Construct the perpendicular from $O$ to the given line.
The foot $T_1$ is its point of contact because a tangent is perpendicular to its contact radius.

The contact $T_1$ lies off line $OP$ .
If it lay on $OP$ , the given line through $T_1$ and $P$ would be $OP$ , a secant through the center.

Construct the perpendicular from $T_1$ to line $OP$ .
From its foot on $OP$ , copy the distance to $T_1$ along the perpendicular on the opposite side to locate $T_2$ .

Reflection fixes $O$ and $P$ .
It carries the circle onto itself and carries the given tangent to the line through $P$ and $T_2$ .
That line has just one contact with the circle.

The points $T_1$ and $T_2$ differ because they lie on opposite sides of $OP$ .
A tangent has just one contact, so these are different lines.

Each contact is the foot of the perpendicular from O to its tangent line.

Answer

Drop a perpendicular from $O$ to the given line to find $T_1$ . Reflect $T_1$ across $OP$ to obtain $T_2$ , then draw line $PT_2$ .

Key idea

A perpendicular recovers the hidden contact before reflection gives the other one.
Problem 7 Place an anchor for two equal cable runs

A circular rim has center $O$ and radius $11$ cm. An anchor $P$ must be placed on a specified ray from $O$ outside the rim.

Two straight cables from the anchor are to touch the rim at distinct points, and each segment from the anchor to its contact point must be $60$ cm long.

Find the required distance $OP$ . Once the anchor is marked, explain how to locate both cable contact points.
1. Hint 1
  At a point where a cable touches, its line is perpendicular to the radius.
2. Hint 2
  The two given lengths are legs of the contact triangle. Once you know the hypotenuse, use it as a diameter.
Answer

Place the anchor $61$ cm from $O$ along the specified ray. The two intersections of the rim and the circle with diameter $OP$ are the cable contact points.

Full solution

A radius and a tangent cable meet at a right angle, so the distance $OP$ is the hypotenuse.

Use the two given leg lengths.
$\begin{aligned}OP^2&=11^2+60^2\\&=3721\\OP&=61\text{ cm}.\end{aligned}$

Mark $P$ that far from $O$ on the specified ray.
This is outside the rim because $61$ cm exceeds $11$ cm.

Bisect $OP$ and use its midpoint as the center of a circle passing through $O$ and $P$ .
Mark both intersections with the rim and join each to $P$ .

Each resulting cable segment has the same hypotenuse and contact radius in its right triangle, so each length is $60$ cm.

The two cable segments follow the two constructed tangent lines.

Answer

Place the anchor $61$ cm from $O$ along the specified ray. The two intersections of the rim and the circle with diameter $OP$ are the cable contact points.

Key idea

Use the contact triangle to place the anchor before constructing the lines.
Problem 8 Find the bisector of the shared chord

For a positive-radius circle centered at $O$ and an outside point $P$ , construct the auxiliary circle with center $M$ and diameter $OP$ .

Its two crossings with the given circle are $T_1$ and $T_2$ . Let $H$ be where line $OP$ meets chord $T_1T_2$ .

Prove that $H$ bisects the chord and that the chord is perpendicular to $OP$ .
1. Hint 1
  The endpoints of the chord lie on both circles. Compare their distances from each center.
2. Hint 2
  Any two distinct points equally far from the endpoints of a segment determine its perpendicular bisector.
Answer

Line $OP$ is the perpendicular bisector of $T_1T_2$ . Its crossing $H$ is the chord's midpoint.

Full solution

The distances from $O$ to $T_1$ and $T_2$ are equal because they are radii of the given circle.

The distances from $M$ to $T_1$ and $T_2$ are equal because they are radii of the auxiliary circle.

Thus both $O$ and $M$ lie on the perpendicular bisector of $T_1T_2$ .
The centers are distinct and lie on $OP$ , so line $OP$ is that bisector.

Its intersection $H$ with the chord therefore bisects it and makes a right angle there.

The same chord belongs to both circles.

Answer

Line $OP$ is the perpendicular bisector of $T_1T_2$ . Its crossing $H$ is the chord's midpoint.

Key idea

Both centers are equally far from the two shared endpoints.
Problem 9 Locate the chord when the circles match in size

A circle centered at $O$ has radius $16$ mm. Point $P$ is $32$ mm from $O$ . The auxiliary circle with diameter $OP$ has center $M$ and crosses the given circle at $T_1$ and $T_2$ .

Line $T_1T_2$ meets $OP$ at $H$ . Find $OH$ and prove that your value follows from the two circles' sizes.
1. Hint 1
  Compute the auxiliary radius and compare it with the original radius.
2. Hint 2
  Consider which segment has endpoints $O$ and $M$ . Ask whether each contact point is equally far from those two endpoints.
Answer

$OH=8$ mm.

Full solution

The auxiliary radius is half of $32$ mm, which is $16$ mm.
The two circles have equal radii.

Each of $T_1$ and $T_2$ is $16$ mm from both $O$ and $M$ .
They therefore lie on the perpendicular bisector of $OM$ .

Line $T_1T_2$ is that perpendicular bisector.
Its crossing $H$ with $OM$ is the midpoint of $OM$ .

Since $OM$ is $16$ mm, $OH$ is $8$ mm.

The common chord is the perpendicular bisector of OM in this special case.

Answer

$OH=8$ mm.

Key idea

Equal radii put both contacts on the perpendicular bisector of the segment joining the centers.
Problem 10 Critique a circle that passes through both endpoints

A circle of positive radius has center $O$ , and $P$ is outside it. The midpoint of $OP$ is $M$ .

A student chooses a point $N$ on the perpendicular bisector of $OP$ , different from $M$ , and draws a circle centered at $N$ through $O$ . The student says the construction works because this circle also passes through $P$ .

Suppose this new circle crosses the original circle at $X$ . Can line $PX$ be tangent at $X$ ? Explain why passing through both endpoints is insufficient and describe the correction.
1. Hint 1
  The semicircle theorem uses a diameter. Where must a diameter's midpoint be?
2. Hint 2
  If line $PX$ were tangent, use the right angle at $X$ to put $X$ on the correct auxiliary circle.
3. Hint 3
  Two distinct circles that already share $O$ and $P$ cannot share a third point. Check that $X$ differs from both endpoints.
Answer

Line $PX$ cannot be tangent at $X$ . Use the circle centered at $M$ with radius $MO$ so $OP$ is a diameter.

Full solution

The distances from $N$ to $O$ and $P$ are equal, so the student's circle really does pass through both points.
Its center is off $OP$ , so $OP$ is only a chord of that circle.

If line $PX$ were tangent at $X$ , its angle with radius $OX$ would be right.
The semicircle converse would put $X$ on the circle with diameter $OP$ .

The circle centered at $N$ and the diameter circle centered at $M$ are distinct.
They already share $O$ and $P$ .
Three common points would give a triangle with two different circumcenters, which is impossible.

Since $O$ is the original circle's center and $P$ is outside it, neither can be $X$ .
Thus $X$ cannot be on the correct diameter circle or be a point of tangency.

Center the compass at $M$ and open it to $O$ .
Now $OP$ is a diameter, and the crossings give the required right angles.

The dotted and dashed circles share O and P. Only the dashed one has OP as a diameter.

Answer

Line $PX$ cannot be tangent at $X$ . Use the circle centered at $M$ with radius $MO$ so $OP$ is a diameter.

Key idea

The center of the auxiliary circle must be the midpoint of the diameter.

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