Geometry · G.C.A.3 · G.C.A.4

Constructions with circles

With only a compass and a straightedge, how do you draw the circle through the three corners of a triangle, the circle inside it that touches all three sides, and the lines from an outside point that just touch a circle?

You need

The inscribed angle theoremGeometry Tangent propertiesGeometry Bisecting a segment and an angleGeometry

Where you are 8 in Compass and straightedge constructions

Geometry11

Compass and straightedge constructions8

▴ 3 above Perpendicular and parallel lines through a point Why a construction works

Constructions with circles

Equilateral triangle, square and hexagon in a circle Dividing a segment into equal parts

When you reach for it

Three situations call for a circle whose center nobody has marked:

You have three points, the corners of a triangle, and you want the one circle that passes through all three. It is the circumscribed circle of the triangle, and its center is the circumcenter.
You have a triangle and you want the largest circle that fits inside it, touching each side once. It is the inscribed circle, and its center is the incenter.
You have a circle and a point outside it, and you want the lines through the point that touch the circle at exactly one point each. They are the tangents from the point.

A compass draws a circle about a center you already have; it cannot find one. So in each case the work is to locate a point exactly, using only the moves of bisecting a segment and an angle, and then the compass does the rest. The same construction finds the center of anything round from a piece of its rim, such as a broken plate (Example 6).

All three rest on one idea: a point fixed by equal distances. A point the same distance from two points lies on a perpendicular bisector; a point the same distance from two lines lies on an angle bisector; and a point where a line meets a radius at a right angle is a point of tangency.

The procedure

The circle through three corners

The construction rests on one fact, the perpendicular bisector theorem:

A point is the same distance from A and from B exactly when it lies on the perpendicular bisector of \overline{AB}.

Construct the perpendicular bisectors of two sides, say \overline{AB} and \overline{AC}. Each one is the segment construction: equal arcs from both ends, and the line through their two crossings.

Mark the point O where the two bisectors cross. They do cross. The sides \overline{AB} and \overline{AC} meet at A, so they are not parallel, and two lines perpendicular to two non-parallel lines are not parallel either. Now O is on the bisector of \overline{AB}, so OA = OB, and it is on the bisector of \overline{AC}, so OA = OC. Together:

OA = OB = OC

With the point on O and the pencil on A, draw the circle. Its radius is OA, and OB and OC are the same length, so the circle passes through B and C too.

Figure 1Two perpendicular bisectors cross at O, which is the same distance from all three corners. The circle about O through A passes through B and C as well.

Why two bisectors are enough. From OB = OA = OC, the point O is the same distance from B and C, so it is already on the bisector of \overline{BC}. Drawing the third bisector adds no new point, but it is a good test of the drawing: three lines should pass through one point.

Why there is only one such circle. The center of any circle through A, B and C is the same distance from all three, so it lies on both bisectors, and two crossing lines meet once. The center must be O and the radius must be OA.

Where the center lands. It depends on the largest angle of the triangle. In a triangle whose angles are all acute, O is inside. In a right triangle, O is the midpoint of the hypotenuse, because the right angle is an inscribed angle on a diameter and so the hypotenuse is a diameter. In a triangle with an obtuse angle, O is outside, beyond the longest side, so the bisectors have to be drawn long enough to meet (Example 2).

The circle inside a triangle

The fact this time is about lines, not points. The distance from a point to a line is the length of the perpendicular from the point to the line, which is the shortest segment from the point to the line. Then, by the angle bisector distance theorem:

A point inside an angle is the same distance from both sides exactly when it lies on the bisector of the angle.

Bisect two of the angles, say at A and at B. Each is the angle construction: an arc about the vertex crossing both sides, then equal arcs from those two crossings.

Mark the point I where the two bisectors cross. Both bisectors run into the inside of the triangle, so they cross inside it. I is on the bisector at A, so it is the same distance from \overline{AB} and \overline{AC}. It is on the bisector at B, so it is the same distance from \overline{AB} and \overline{BC}. So I is one distance, call it r, from all three sides.

Drop the perpendicular from I to one side, say \overline{AB}, and call its foot T. With the point on I, swing an arc that crosses \overline{AB} twice, at X and Y. Then IX = IY, so I lies on the perpendicular bisector of \overline{XY}. Bisect \overline{XY}: the line through I and the midpoint of \overline{XY} is perpendicular to \overline{AB}, and it meets \overline{AB} at T. This is the perpendicular from a point off a line.

With the point on I and the pencil on T, draw the circle. Its radius is IT = r.

Figure 2The angle bisectors at A and B cross at I, the same distance from all three sides. That distance is the perpendicular IT, and the circle with radius IT touches each side once.

Why the circle touches each side once. On each side, the foot of the perpendicular from I is the distance r from I, so it is on the circle. Take any other point Q of that side. The triangle ITQ has a right angle at T, so IQ is its hypotenuse and IQ > IT = r: the point Q is outside the circle. So each side meets the circle at its foot and nowhere else, which makes each side a tangent, perpendicular to the radius where it touches, as in tangent properties.

Why the foot has to be constructed. The radius is the perpendicular distance from I to a side. The distance from I to a corner is longer, and the distance to the midpoint of a side is longer unless the foot happens to be the midpoint. A circle drawn through either one crosses the sides instead of touching them.

A tangent from a point outside a circle

Let the circle have center O and radius r, and let P be outside it. A line through a point T of the circle is a tangent exactly when it is perpendicular to the radius \overline{OT}. So the task is to find a point T on the circle with \angle OTP = 90^\circ. The points that see a segment at a right angle are the points of the circle that has the segment as a diameter. That is the plan.

Bisect \overline{OP} and call its midpoint M.

With the point on M and the pencil on O, draw a second circle. Since MP = MO, it passes through P as well: \overline{OP} is its diameter.

Mark T and U, the two points where the second circle crosses the given one. It does cross it. The second circle passes through O, which is inside the given circle, and through P, which is outside, so each of its two halves on either side of \overleftrightarrow{OP} crosses the given circle once. The two crossings are mirror images in \overleftrightarrow{OP}.

Draw \overleftrightarrow{PT} and \overleftrightarrow{PU}. These are the two tangents.

Figure 3The circle on the diameter \overline{OP} crosses the given circle at T and U. The angle OTP is inscribed on a diameter, so it is a right angle, and \overleftrightarrow{PT} is a tangent.

Why it works. T is on the circle with diameter \overline{OP}, so \angle OTP is an inscribed angle on a diameter and measures 90^\circ. So \overleftrightarrow{PT} is perpendicular to the radius \overline{OT} at T. Any other point Q of that line makes a right triangle OTQ with hypotenuse OQ > OT = r, so Q is outside the given circle. The line meets the circle only at T: it is a tangent. The same holds at U.

The tangent pieces are equal. The triangles OTP and OUP have right angles at T and U, share the hypotenuse \overline{OP}, and have equal legs OT = OU = r, so they are congruent by hypotenuse leg. In particular

PT = PU = \sqrt{OP^2 - r^2}

by the Pythagorean theorem.

Euclid gives the two triangle constructions in Book IV of the Elements: the circle inside a triangle in Proposition 4 and the circle through its corners in Proposition 5.

Worked examples

Example 1

A circumscribed circle on a grid

Triangle ABC has corners A = (0, 0), B = (8, 0) and C = (2, 6). The perpendicular bisectors of \overline{AB} and \overline{AC} are constructed. Find where they cross and the radius of the circumscribed circle, and confirm that the bisector of \overline{BC} passes through the same point.

Solution

\overline{AB} lies along the x-axis with midpoint (4, 0), so its perpendicular bisector is the vertical line x = 4.

\overline{AC} has midpoint (1, 3) and slope \displaystyle \frac{6}{2} = 3. Its bisector is perpendicular to it, so its slope is \displaystyle -\frac{1}{3}, and it passes through the midpoint:

y - 3 = -\tfrac{1}{3}(x - 1)

At x = 4 this gives y = 3 - 1 = 2. So the bisectors cross at O = (4, 2).

The radius is the distance from O to any corner. By the distance formula,

\begin{aligned} OA &= \sqrt{4^2 + 2^2} = \sqrt{20} = 2\sqrt{5} \approx 4.47 \\ OB &= \sqrt{(8 - 4)^2 + (0 - 2)^2} = \sqrt{20} \\ OC &= \sqrt{(2 - 4)^2 + (6 - 2)^2} = \sqrt{20} \end{aligned}

All three agree, as the construction promises.

Check with the third bisector. \overline{BC} has midpoint (5, 3) and slope \displaystyle \frac{6 - 0}{2 - 8} = -1, so its bisector has slope 1: y - 3 = x - 5, that is y = x - 2. At x = 4, y = 2, so it passes through O. The angles of this triangle are all acute, and O lies inside it.

Example 2

When the center falls outside

Triangle ABC has corners A = (0, 0), B = (10, 0) and C = (2, 2). Find the center and radius of its circumscribed circle, and say where the center lies.

Solution

The bisector of \overline{AB} is x = 5. \overline{AC} has midpoint (1, 1) and slope 1, so its bisector has slope -1:

y - 1 = -(x - 1), \quad\text{that is,}\quad y = -x + 2

At x = 5, y = -3. So O = (5, -3), and

\begin{aligned} OA &= \sqrt{5^2 + 3^2} = \sqrt{34} \approx 5.83 \\ OC &= \sqrt{3^2 + 5^2} = \sqrt{34} \end{aligned}

O is below the x-axis, while the whole triangle is on or above it. The center is outside the triangle, on the far side of \overline{AB} from C.

Figure 4Example 2. The angle at C is obtuse, so the bisectors cross outside the triangle, on the far side of the longest side \overline{AB}.

That placement says the angle at C is obtuse. The chord \overline{AB} cuts the circle into two arcs, and the center is on the side of the arc away from C, so C sits on the minor arc. The inscribed angle \angle ACB opens onto the other arc, the major one, which measures more than 180^\circ, and an inscribed angle is half its arc, so m\angle ACB > 90^\circ.

For the drawing, this means the two bisectors must be extended well past \overline{AB} before they meet. A short pencil line drawn only across the triangle never shows the center.

Example 3

An inscribed circle

Triangle ABC has corners A = (0, 0), B = (17, 0) and C = (7, 24). Its angle bisectors at A and B are constructed. Show that they cross at I = (8, 6), find the radius of the inscribed circle, and find where the circle touches \overline{AB}.

Solution

First the side lengths: \displaystyle AC = \sqrt{7^2 + 24^2} = 25 and \displaystyle BC = \sqrt{10^2 + 24^2} = 26, with AB = 17.

The bisector at A. An arc of radius 25 about A crosses line AB at (25, 0), past B, and crosses \overline{AC} at C itself. A is the same distance from these two points, and so is every point of the bisector, so the bisector passes through their midpoint, \displaystyle \left(\frac{25 + 7}{2}, \frac{0 + 24}{2}\right) = (16, 12). It is the line through (0, 0) and (16, 12), which is \displaystyle y = \frac{3}{4}x. At x = 8, y = 6, so I is on it.

The bisector at B. An arc of radius 26 about B crosses line AB at (-9, 0), past A, and crosses \overline{BC} at C. Their midpoint is (-1, 12), and the line through B = (17, 0) and (-1, 12) has slope \displaystyle \frac{12}{-18} = -\frac{2}{3}:

y = -\tfrac{2}{3}(x - 17)

At x = 8, \displaystyle y = -\frac{2}{3}(-9) = 6, so I is on this bisector too. The two cross at I = (8, 6).

The radius. \overline{AB} lies along the x-axis, so the perpendicular from I drops straight down to T = (8, 0). In the construction, an arc of radius 7.5 about I crosses the axis where (x - 8)^2 + 6^2 = 56.25, at (3.5, 0) and (12.5, 0), whose midpoint is T. The radius is r = IT = 6, and the circle touches \overline{AB} at (8, 0), which is 8 from A.

Check on a second side. The two tangent pieces from A are equal, so the circle should touch \overline{AC} at the point F that is 8 from A, which is \displaystyle \frac{8}{25} of the way to C: \displaystyle F = \left(\frac{56}{25}, \frac{192}{25}\right). Then

\begin{aligned} IF &= \sqrt{\left(\tfrac{144}{25}\right)^2 + \left(\tfrac{42}{25}\right)^2} \\ &= \tfrac{150}{25} = 6 \end{aligned}

and the slope of \overline{IF} is \displaystyle -\frac{7}{24} while the slope of \overline{AC} is \displaystyle \frac{24}{7}. Their product is -1, so \overline{IF} is perpendicular to \overline{AC} and the distance from I to that side is 6 as well.

Example 4

Tangents in coordinates

A circle has center O = (0, 0) and radius 5, and P = (13, 0). Find the midpoint M of \overline{OP}, the points T and U where the circle on \overline{OP} crosses the given circle, and the length of the tangent \overline{PT}.

Solution

M = (6.5, 0), and the second circle has center M and radius MO = 6.5. Using the equation of a circle for each:

\begin{aligned} x^2 + y^2 &= 25 \\ (x - 6.5)^2 + y^2 &= 42.25 \end{aligned}

Subtracting the second equation from the first removes y^2. Since x^2 - (x - 6.5)^2 = 13x - 42.25, this leaves 13x - 42.25 = -17.25, so 13x = 25 and \displaystyle x = \frac{25}{13}. Then

\begin{aligned} y^2 &= 25 - \tfrac{625}{169} = \tfrac{3600}{169} \\ y &= \pm\tfrac{60}{13} \end{aligned}

So \displaystyle T = \left(\frac{25}{13}, \frac{60}{13}\right) and \displaystyle U = \left(\frac{25}{13}, -\frac{60}{13}\right).

The tangent length. From T to P the run is \displaystyle 13 - \frac{25}{13} = \frac{144}{13} and the rise is \displaystyle \frac{60}{13}, so

\begin{aligned} PT &= \sqrt{\left(\tfrac{144}{13}\right)^2 + \left(\tfrac{60}{13}\right)^2} \\ &= \sqrt{144} = 12 \end{aligned}

which matches \displaystyle \sqrt{OP^2 - r^2} = \sqrt{13^2 - 5^2} = 12.

Check the right angle. The slope of \overline{OT} is \displaystyle \frac{60}{25} = \frac{12}{5}. The slope of \overline{PT} is \displaystyle \frac{60/13}{25/13 - 13} = \frac{60}{-144} = -\frac{5}{12}. Their product is -1, so the radius and the line meet at a right angle at T.

Example 5

The chord between the two points of contact

A circle has radius 6, and a point P is 10 from its center O. The tangents from P touch the circle at T and U. Find PT, the length of \overline{TU}, and the distance from O to the point N where \overline{TU} crosses \overline{OP}.

Solution

The triangle OTP has a right angle at T, so

PT = \sqrt{10^2 - 6^2} = 8

O is the same distance from T and U, and so is P, since the tangent pieces are equal. Two points each equidistant from T and U fix the perpendicular bisector of \overline{TU}, so \overleftrightarrow{OP} is that bisector. It crosses \overline{TU} at right angles at its midpoint N, and TN is the height of triangle OTP measured from T to the side \overline{OP}.

Find that height from the area of the triangle, counted two ways. With the legs as base and height, the area is \displaystyle \frac{1}{2}(6)(8) = 24. With \overline{OP} as base, it is \displaystyle \frac{1}{2}(10)(TN). So

5 \cdot TN = 24, \qquad TN = 4.8

and TU = 2(4.8) = 9.6.

Finally, triangle ONT has a right angle at N and hypotenuse OT = 6:

ON = \sqrt{6^2 - 4.8^2} = \sqrt{12.96} = 3.6

The construction would use a helper circle of radius 5 about the midpoint of \overline{OP}. Nothing above needed it, but it is where T and U come from.

Example 6

The center of a broken plate

A piece of a broken round plate still has part of its rim. On a grid marked in centimeters laid over the piece, three points of the rim are A = (-6, 8), B = (6, 8) and C = (8, 6). Find the center of the whole plate and its diameter.

Solution

The rim is a circle through A, B and C, and there is only one such circle: the circumscribed circle of triangle ABC. So the center is where the perpendicular bisectors of two of the chords cross.

\overline{AB} is level with midpoint (0, 8), so its bisector is x = 0.

\overline{BC} has midpoint (7, 7) and slope \displaystyle \frac{6 - 8}{8 - 6} = -1, so its bisector has slope 1: y - 7 = x - 7, that is y = x.

The two cross at (0, 0), the center. The radius is

OA = \sqrt{6^2 + 8^2} = 10

and \displaystyle OC = \sqrt{8^2 + 6^2} = 10 agrees. The plate is 20 cm across.

On a real plate, choose the three points as far apart along the rim as the piece allows. The perpendicular bisector of a short chord is easily tipped by a small error in the two points, and two bisectors that cross at a shallow angle carry that error a long way.

Boundaries

Three points on one line have no circle. The perpendicular bisectors of two of the segments are both perpendicular to that line, so they are parallel and never meet (Exercise 16). Three points not on one line always have exactly one circle through them.

Four points need not. A fourth point lies on the circle through the first three only in special cases; a quadrilateral whose corners do lie on one circle is a cyclic quadrilateral, and its opposite angles add to 180^\circ.

The circumcenter can be outside, or on a side. It is outside when one angle is obtuse and at the midpoint of the hypotenuse when one angle is right. The construction still works; the lines only have to be long enough.

The incenter is always inside, and every triangle has an inscribed circle. A polygon with more sides need not have one: a rectangle that is not a square has no circle touching all four sides, because its angle bisectors do not all meet at one point.

The radius of the inscribed circle is a perpendicular distance. It has to be measured along the perpendicular from I to a side, never to a corner or to the midpoint of a side.

The tangent construction needs P outside the circle. For P on the circle, the circle on \overline{OP} touches the given circle only at P, and there is exactly one tangent: the line perpendicular to \overline{OP} at P, which is the perpendicular at a point on a line. For P inside, there is no tangent at all (Exercise 20).

Exact in principle. The coordinates in the examples say where the constructions land. None of them is measured during a construction; the straightedge has no marks.

Where this goes next

The two centers found here have lessons of their own. The circumcenter lesson follows the center through acute, right and obtuse triangles and into coordinates, and the incenter lesson works with the tangent pieces from each corner and the radius of the inscribed circle. Both are cases of a locus: the center is the one point that belongs to two loci at once.

The circle through the corners leads to cyclic quadrilaterals and to the regular polygons drawn inside a circle in the equilateral triangle, square and hexagon, where the circle comes first and the polygon is fitted to it. The tangent construction leads back to tangent properties, where the equal tangent pieces and the angle between two tangents are studied for their own sake. And every step here can be defended the way any construction is justified: name the equal lengths the compass made, and read off what they force.

Exercises

The circle through three corners

Triangle ABC has corners A = (0, 0), B = (12, 0) and C = (4, 8). Find the center and radius of its circumscribed circle.
Triangle ABC has corners A = (0, 0), B = (10, 0) and C = (8, 4). Find the center and radius of its circumscribed circle. What does the position of the center say about the angle at C?
Triangle ABC has corners A = (0, 0), B = (8, 0) and C = (1, 1). Find the center and radius of its circumscribed circle, and say where the center lies.
A right triangle has legs of 9 cm and 12 cm. Where do the perpendicular bisectors of its sides cross, and what is the radius of its circumscribed circle?
An equilateral triangle has sides of length 6. Find the radius of its circumscribed circle.

The circle inside a triangle

Triangle ABC has corners A = (0, 0), B = (12, 0) and C = (0, 9). Find the incenter I, the radius of the inscribed circle, and the point where it touches \overline{BC}.
Triangle ABC has corners A = (0, 0), B = (6, 0) and C = (3, 4). Find I, the radius of the inscribed circle, and the point where it touches \overline{AC}.
Find the radius of the inscribed circle of an equilateral triangle with sides of length 6, and explain why its center is the center of the circumscribed circle too.
In triangle ABC, m\angle A = 50^\circ and m\angle B = 70^\circ, and the bisectors of these two angles cross at I. Find m\angle AIB.

Tangents from a point outside

A circle has radius 8, and P is 17 from its center. What radius does the helper circle on \overline{OP} have, and how long is each tangent from P?
The circle x^2 + y^2 = 25 has center O = (0, 0), and P = (5, 5). Find the points where the tangents from P touch the circle, and the equations of the two tangents.
A circle has radius 6, and P is 12 from its center. Find the length of each tangent from P and the angle between the two tangents.
A circle has radius 5, and each tangent from P is 12 long. Find OP and the length of the chord \overline{TU} joining the two points of contact.
A circle has center O = (0, 0) and radius \displaystyle \sqrt{5}, and P = (5, 0). Find the two points of contact and the length of each tangent.

Finding a center

Three points on the rim of a broken wheel are A = (1, 7), B = (7, 7) and C = (9, 3). Find the center and radius of the wheel.
Show that there is no circle through (0, 0), (2, 2) and (5, 5) by finding the perpendicular bisectors of the segments from the first point to the second and from the second to the third.

Reasoning

Explain why, once the perpendicular bisectors of \overline{AB} and \overline{BC} cross at O, the bisector of \overline{AC} must pass through O as well.
In the construction of the inscribed circle, prove that side \overline{AB} meets the circle only at T.
In the tangent construction, explain why \angle OTP is a right angle and why that makes \overleftrightarrow{PT} a tangent.
Explain why no tangent can be drawn to a circle from a point inside it, and why a point on the circle has exactly one tangent.
Explain why bisecting two angles of a triangle is enough to find the incenter: why must the third angle bisector pass through the same point?
Prove that the two tangents from an outside point P have equal lengths, and that \overleftrightarrow{OP} bisects the angle between them.

Answers

Center (6, 2), radius \displaystyle 2\sqrt{10} \approx 6.32. The bisector of \overline{AB} is x = 6. \overline{AC} has midpoint (2, 4) and slope 2, so its bisector is \displaystyle y - 4 = -\frac{1}{2}(x - 2), which gives y = 2 at x = 6. Then \displaystyle OA = \sqrt{6^2 + 2^2} = \sqrt{40} and \displaystyle OC = \sqrt{(-2)^2 + 6^2} = \sqrt{40}.
Center (5, 0), radius 5. The bisector of \overline{AB} is x = 5. \overline{AC} has midpoint (4, 2) and slope \displaystyle \frac{1}{2}, so its bisector is y = -2x + 10, which gives y = 0 at x = 5. The center is the midpoint of \overline{AB}, so \overline{AB} is a diameter and the angle at C is a right angle. Check: \displaystyle OC = \sqrt{3^2 + 4^2} = 5.
Center (4, -3), radius 5, outside the triangle, on the far side of \overline{AB} from C. The bisector of \overline{AB} is x = 4. \overline{AC} has midpoint \displaystyle \left(\frac{1}{2}, \frac{1}{2}\right) and slope 1, so its bisector is y = -x + 1, which gives y = -3 at x = 4. \displaystyle OA = \sqrt{4^2 + 3^2} = 5 and \displaystyle OC = \sqrt{3^2 + 4^2} = 5. The angle at C is obtuse.
At the midpoint of the hypotenuse; radius 7.5 cm. The hypotenuse is \displaystyle \sqrt{9^2 + 12^2} = 15 cm, and the right angle is an inscribed angle on a diameter, so the hypotenuse is a diameter.
\displaystyle 2\sqrt{3} \approx 3.46. Let M be the midpoint of \overline{AB}. The bisector of \overline{AB} passes through C, since CA = CB, and \displaystyle CM = \sqrt{6^2 - 3^2} = 3\sqrt{3}. With OM = x, the conditions OA = OC give \displaystyle 9 + x^2 = (3\sqrt{3} - x)^2 = 27 - 6\sqrt{3}x + x^2, so \displaystyle x = \sqrt{3} and \displaystyle OA = \sqrt{9 + 3} = 2\sqrt{3}.
I = (3, 3), radius 3, touching \overline{BC} at (4.8, 5.4). The angle at A is right with the axes as sides, so its bisector is y = x. \displaystyle BC = \sqrt{12^2 + 9^2} = 15. The bisector at B runs through B and the midpoint (1.2, 3.6) of A and the point (2.4, 7.2) of \overline{BC} that is 12 from B, so it is \displaystyle y = -\frac{1}{3}(x - 12), and at x = 3, y = 3. I is 3 from both axes. Side \overline{BC} is 3x + 4y = 36; moving from I along the perpendicular direction (3, 4) gives 3(3 + 3t) + 4(3 + 4t) = 36, so \displaystyle t = \frac{3}{5} and the foot is (4.8, 5.4), at distance 5t = 3 from I.
I = (3, 1.5), radius 1.5, touching \overline{AC} at (1.8, 2.4). CA = CB = 5, so the bisector at C is the line of symmetry x = 3. The bisector at A runs through the midpoint (4, 2) of (5, 0) and (3, 4), both 5 from A, so it is \displaystyle y = \frac{1}{2}x, and at x = 3, y = 1.5. The radius is the distance to \overline{AB}, 1.5. Line AC is \displaystyle y = \frac{4}{3}x; from I along (-4, 3), the point (3 - 4t, 1.5 + 3t) is on it when \displaystyle 1.5 + 3t = \frac{4}{3}(3 - 4t), so \displaystyle t = \frac{3}{10}, giving (1.8, 2.4) at distance 5t = 1.5.
\displaystyle \sqrt{3} \approx 1.73. In an equilateral triangle the bisector of the angle at C is also the perpendicular bisector of \overline{AB}, because CA = CB makes the triangle symmetric about it. The same holds at each corner, so the angle bisectors and the perpendicular bisectors are the same three lines and cross at the same point. The radius is the distance OM from Exercise 5, which is \displaystyle \sqrt{3}, half the circumradius.
m\angle AIB = 120^\circ. The bisectors give m\angle IAB = 25^\circ and m\angle IBA = 35^\circ, and the angles of triangle AIB add to 180^\circ: 180^\circ - 25^\circ - 35^\circ = 120^\circ.
Helper radius 8.5; tangent length 15. The helper circle has diameter OP = 17, and \displaystyle PT = \sqrt{17^2 - 8^2} = \sqrt{225} = 15.
Points of contact (5, 0) and (0, 5); tangents x = 5 and y = 5. The helper circle has center (2.5, 2.5) and r^2 = 12.5, so (x - 2.5)^2 + (y - 2.5)^2 = 12.5, which expands to x^2 + y^2 - 5x - 5y = 0. With x^2 + y^2 = 25 this gives x + y = 5, and then x^2 + (5 - x)^2 = 25 gives x = 0 or x = 5. The radius to (5, 0) is level, so the tangent there is the vertical line x = 5; likewise y = 5 at (0, 5).
Each tangent is \displaystyle 6\sqrt{3} \approx 10.39, and the angle between them is 60^\circ. \displaystyle PT = \sqrt{12^2 - 6^2} = \sqrt{108}. The helper circle has radius 6, so with M the midpoint of \overline{OP}, MT = MO = OT = 6: triangle OMT is equilateral and m\angle TOP = 60^\circ. Then m\angle OPT = 180^\circ - 90^\circ - 60^\circ = 30^\circ, and the two tangents make 2 \cdot 30^\circ = 60^\circ.
OP = 13 and \displaystyle TU = \frac{120}{13} \approx 9.23. \displaystyle OP = \sqrt{5^2 + 12^2}. The area of triangle OTP is \displaystyle \frac{1}{2}(5)(12) = 30 = \frac{1}{2}(13)h, so the height from T to \overline{OP} is \displaystyle h = \frac{60}{13}, and TU = 2h.
(1, 2) and (1, -2); each tangent is \displaystyle 2\sqrt{5} \approx 4.47. The helper circle is (x - 2.5)^2 + y^2 = 6.25, that is x^2 + y^2 - 5x = 0; with x^2 + y^2 = 5 this gives x = 1, and then y^2 = 4. \displaystyle PT = \sqrt{4^2 + 2^2} = \sqrt{20}.
Center (4, 3), radius 5. The bisector of \overline{AB} is x = 4. \overline{BC} has midpoint (8, 5) and slope -2, so its bisector is \displaystyle y - 5 = \frac{1}{2}(x - 8), which gives y = 3 at x = 4. \displaystyle OA = \sqrt{3^2 + 4^2} = 5 and \displaystyle OC = \sqrt{5^2 + 0^2} = 5.
The first segment has midpoint (1, 1) and slope 1, so its bisector is y = -x + 2. The second has midpoint (3.5, 3.5) and slope 1, so its bisector is y = -x + 7. Both have slope -1 and different intercepts, so they are parallel and never meet: no point is the same distance from all three. The three points all lie on y = x.
O on the bisector of \overline{AB} gives OA = OB, and on the bisector of \overline{BC} gives OB = OC. So OA = OC, and a point the same distance from A and C lies on the perpendicular bisector of \overline{AC}.
\overline{IT} is perpendicular to \overline{AB}. For any other point Q of \overline{AB}, triangle ITQ has a right angle at T, so IQ is its hypotenuse and is longer than the leg IT = r. So Q is outside the circle, and T is the only point of \overline{AB} on it.
T lies on the helper circle, whose diameter is \overline{OP}, so \angle OTP is an inscribed angle opening onto a semicircle, half of 180^\circ, which is 90^\circ. So \overleftrightarrow{PT} is perpendicular to the radius \overline{OT} at T. Every other point Q of that line has OQ > OT, as the hypotenuse of right triangle OTQ, so Q is outside the circle and the line meets the circle only at T.
If a line through P touched the circle at T, it would be perpendicular to \overline{OT}, and in the right triangle OTP the hypotenuse OP would be longer than OT = r, putting P outside. So a point inside has no tangent. A point P on the circle has OP = r, so it can only be the point of contact itself, and the one line through P perpendicular to \overline{OP} is the only tangent through it.
I on the bisector at A is the same distance from \overline{AB} and \overline{AC}, and on the bisector at B the same distance from \overline{AB} and \overline{BC}. So it is the same distance from \overline{CA} and \overline{CB}, and since it is inside the angle at C, the angle bisector distance theorem puts it on the bisector of that angle.
Triangles OTP and OUP have right angles at T and U, share the hypotenuse \overline{OP}, and have OT = OU as radii. By hypotenuse leg they are congruent, so PT = PU, and the corresponding angles \angle OPT and \angle OPU are equal, which says \overleftrightarrow{OP} bisects \angle TPU.