Tangent properties
A line touches a circle at a single point. What does that one touch force on the radius, on a second tangent from the same point, and on a polygon or a second circle that the line also touches?
Where you are 15 in Circles
Why this exists
A tangent is a line that touches a circle and does not cut it; chords, secants and tangents sets out the words. Touching at one point sounds like too weak a condition to say much. It turns out to fix a great deal. The radius to the point of contact always meets the tangent at a right angle, and once there is a right angle there is a right triangle, so the Pythagorean theorem and the angle sum of a triangle take over.
That one right angle is behind every result in this lesson. Two tangents drawn from one outside point are equally long. A polygon drawn around a circle has its sides cut into matching pairs of pieces. Two circles can share up to four tangent lines, and their lengths follow from the two radii and the distance between the centers. A belt running round two wheels, a pipe lying in a V-shaped cradle and a coin pushed into the corner of a frame are all these facts at work.
Definition
A line is tangent to a circle when it meets the circle at exactly one point, the point of tangency or point of contact.
From a point P outside a circle, a tangent segment is the piece \overline{PA} of a tangent line running from P to its point of contact A. Its length PA is the tangent length from P. Two tangent segments from the same point, \overline{PA} and \overline{PB}, make the angle \angle APB, called a circumscribed angle of the circle.
A polygon is circumscribed about a circle when every one of its sides is tangent to the circle, and the circle is then inscribed in the polygon. Each side touches the circle at one point, which splits the side into two pieces. For a triangle the inscribed circle is the incircle, centered at the Incenter.
A line tangent to two circles at once is a common tangent of the two. It is an external common tangent when both circles lie on the same side of it, and an internal common tangent when they lie on opposite sides, so that it crosses the segment joining the centers. The length of a common tangent is the distance between its two points of contact.
Intuition
Look at Figure 1 as if the line OP were a mirror. Reflecting in it swaps the two halves of the picture. The circle, centered on the mirror, lands on itself, and P, on the mirror, stays put. So the tangent from P touching at A lands on a line from P that still touches the circle once: the tangent touching at B. Whatever is true of one tangent segment is true of the other, and in particular PA = PB. The four points O, A, P, B make a kite with OP as its line of symmetry.
A polygon around a circle is a ring of such pictures. Each corner is an outside point with two tangents running from it along the two sides that meet there, so each corner owns two equal pieces. Label the pieces by their corner, as in Figure 2, and every side is the sum of two different letters. Adding a pair of opposite sides then uses each letter exactly once, whichever pair you take.
For two circles, picture sliding one toward the other. Far apart, there is room for four common tangents: two running along the outside of the pair and two crossing between them. When the circles touch, the two crossing tangents close up into the single line through the point of contact. When they overlap, no line can pass between them, and only the two outside tangents are left. When the small circle sits inside the large one and touches it, one tangent remains, and once it is fully inside there are none.
Results
Tangent and radius. A tangent is perpendicular to the radius drawn to its point of contact. Conversely, the line through a point A of the circle perpendicular to the radius OA is tangent at A.
Two tangents from an outside point. If \overline{PA} and \overline{PB} are tangent at A and B to a circle with center O, then
and the line OP bisects both \angle APB and \angle AOB and is the perpendicular bisector of the chord \overline{AB}.
The circumscribed angle. In the same figure, the angle at P and the central angle at O are supplementary:
so m\angle APB = 180^\circ - m\overset{\frown}{AB}, where \overset{\frown}{AB} is the minor arc.
Circumscribed polygons. The two pieces at each corner of a circumscribed polygon are equal. For a circumscribed quadrilateral ABCD the two pairs of opposite sides have equal sums:
For a triangle the pieces at the corners are s - a, s - b and s - c, where s is half the perimeter, as worked out in the incenter lesson. A polygon with perimeter p circumscribed about a circle of radius r has area
Common tangents. Take two different circles with radii R \ge r whose centers are d apart. The number of common tangents depends only on how d compares with R + r and R - r:
| Circles | Condition | Tangents |
|---|---|---|
| apart | d > R + r | 4 |
| touch outside | d = R + r | 3 |
| cross | R - r < d < R + r | 2 |
| touch inside | d = R - r | 1 |
| one inside | d < R - r | 0 |
The length of a common tangent, between its two points of contact, is
for an external and an internal tangent. A tangent of that kind exists exactly when the number under the root is not negative.
Why it is true
Tangent and radius. Every point of a tangent other than A lies outside the circle, so it is farther from O than A is. That makes \overline{OA} the shortest segment from O to the line, and the shortest segment from a point to a line is the perpendicular one. The full argument, in both directions, is proving a tangent is perpendicular to the radius.
Two tangents. Join O to A, B and P. The triangles OAP and OBP have right angles at A and B by the first result, equal legs OA = OB because both are radii, and the hypotenuse OP in common. By hypotenuse leg congruence they are congruent, so PA = PB, \angle APO \cong \angle BPO and \angle AOP \cong \angle BOP. Then P is the same distance from A as from B, and so is O. Every point equally far from the two ends of a segment lies on its perpendicular bisector, so the line OP is the perpendicular bisector of \overline{AB}.
The circumscribed angle. OAPB is a quadrilateral, so its angles add to 360^\circ (angle sums of polygons). The angles at A and B are right angles, which leaves 180^\circ for the angles at P and O. The central angle \angle AOB has the same measure as the minor arc \overset{\frown}{AB}, by the definition of arc measure.
Opposite sides of a circumscribed quadrilateral. Name the pieces by their corners as in Figure 2: w at A, x at B, y at C and z at D. Then
and both are w + x + y + z.
Area. Join the center O to every corner. The circle lies inside the polygon and O inside the circle, so this cuts the polygon into triangles without overlap, one standing on each side with its apex at O. The radius to the point of contact is perpendicular to that side, so each triangle has height r. Adding \tfrac{1}{2} r \times \text{side} over all the sides gives \tfrac{1}{2} r p.
Length of a common tangent.
Let an external tangent touch the circles at A and B, with R \ge r. Both radii O_1A and O_2B are perpendicular to the tangent, so they are parallel and point to the same side of it. The line through O_2 parallel to AB meets \overline{O_1A} at C. Then CABO_2 is a rectangle, so CO_2 = AB and CA = r, which leaves O_1C = R - r. The triangle O_1CO_2 has a right angle at C and hypotenuse d, so
For an internal tangent the two radii point to opposite sides of the tangent, C lands on O_1A extended past A, and the leg becomes O_1C = R + r.
The same triangle explains the count. A right triangle with hypotenuse d and a leg R - r can be built only when d \ge R - r. When d is larger it can be built on either side of the line of centers, giving two external tangents; when d = R - r it flattens into a segment and the two become one. The internal tangents behave the same way with R + r in place of R - r. Adding the two counts gives the table.
Worked examples
A circumscribed angle from its arc
Two tangents from P touch a circle with center O at A and B, and the minor arc \overset{\frown}{AB} measures 124^\circ, as in Figure 5. Find m\angle APB, m\angle PAB and m\angle OAB.
The central angle \angle AOB cuts off the minor arc, so m\angle AOB = 124^\circ. The angles at A and B in the quadrilateral OAPB are right angles, so the circumscribed angle is the supplement of the central angle:
The tangent segments PA and PB are equal, so \triangle PAB is isosceles and its two base angles are equal. They share what is left of 180^\circ:
The radii OA and OB are equal too, so \triangle OAB is isosceles with apex angle 124^\circ:
Check. The chord \overline{AB} splits the right angle \angle OAP into these two pieces, and 62^\circ + 28^\circ = 90^\circ. Notice also that 62^\circ is half the arc of 124^\circ. The angle between a tangent and a chord is always half the arc it cuts off, one of the results in angles made by chords, secants and tangents.
Tangent lengths written as expressions
A circle has center O and radius 21. The two tangent segments from an outside point P touch it at A and B, with PA = 3x + 2 and PB = 5x - 10. Find x, the tangent length, the distance OP, and how far P is from the nearest point of the circle.
SolutionTwo tangent segments from one point are equal, so set the expressions equal:
The tangent length is PA = 3(6) + 2 = 20, and PB = 5(6) - 10 = 20 agrees.
The radius OA is perpendicular to the tangent at A, so \triangle OAP has a right angle at A, legs 21 and 20, and hypotenuse OP:
so OP = 29.
The segment \overline{OP} crosses the circle at a point M with OM = 21, so PM = 29 - 21 = 8. No point X of the circle is nearer: by the triangle inequality, OP \le OX + XP, so XP \ge 29 - 21 = 8. The nearest point of the circle is 8 from P.
Tangent pieces around a quadrilateral
Quadrilateral ABCD is circumscribed about a circle that touches \overline{AB} at P, \overline{BC} at Q, \overline{CD} at R and \overline{DA} at S, lettered as in Figure 2. The sides are AB = 11, BC = 14 and CD = 17, and AP = 4. Find all eight tangent pieces, the side DA and the perimeter.
SolutionWork round the quadrilateral. At each corner the two pieces are equal, and on each side the two pieces add to the side.
- At A: AS = AP = 4.
- On \overline{AB}: PB = 11 - 4 = 7, so at B, BQ = 7 as well.
- On \overline{BC}: QC = 14 - 7 = 7, so at C, CR = 7.
- On \overline{CD}: RD = 17 - 7 = 10, so at D, DS = 10.
The last side is made of the two pieces left over: DA = DS + SA = 10 + 4 = 14. The perimeter is
and that is twice the sum of the four different pieces, 2(4 + 7 + 7 + 10) = 56, since each corner’s piece appears once on each of its two sides.
Check. Opposite sides should have equal sums: AB + CD = 11 + 17 = 28 and BC + DA = 14 + 14 = 28. They do. The perimeter could have been found without the pieces at all, as 2(AB + CD) = 56.
A regular hexagon around a circle
A regular hexagon is circumscribed about a circle of radius 6. Find the side of the hexagon and its area, and compare the area with the circle’s.
SolutionEach corner has two equal pieces, and here all six sides are equal too, so every piece is half a side and each point of contact is the midpoint of its side.
Join the center O to the six corners. That cuts the hexagon into six triangles, each standing on one side. The radius to the point of contact is perpendicular to that side and meets it at its midpoint, so each triangle is two right triangles with legs 6 and half a side. The six triangles are therefore congruent and isosceles. Their apex angles fill the full turn at O, so each is 360^\circ \div 6 = 60^\circ, and an isosceles triangle with a 60^\circ apex has 60^\circ base angles too. Each triangle is equilateral.
An equilateral triangle with side s has height \displaystyle \frac{\sqrt{3}}{2}s, and here the height is the radius:
The perimeter is \displaystyle 6 \cdot 4\sqrt{3} = 24\sqrt{3}, and the area of a circumscribed polygon is half the radius times the perimeter:
The circle has area \pi \cdot 6^2 = 36\pi \approx 113.1. The hexagon is larger by about 11.6, as it must be, since the circle lies inside it. The radius of the inscribed circle is the hexagon’s The apothem of a regular polygon, and this calculation is the area of a regular polygon seen from the circle.
Counting common tangents
Two circles have radii 5 and 3. How many common tangents do they have when their centers are 10, 8, 5, 2 and 1 apart? For the first distance, find the lengths of the tangents too.
SolutionEverything is decided by comparing d with R + r = 8 and R - r = 2.
| d | Comparison | Circles | Tangents |
|---|---|---|---|
| 10 | 10 > 8 | apart | 4 |
| 8 | 8 = 8 | touch outside | 3 |
| 5 | 2 < 5 < 8 | cross | 2 |
| 2 | 2 = 2 | touch inside | 1 |
| 1 | 1 < 2 | one inside | 0 |
At d = 10 the two external tangents each have length \displaystyle \sqrt{10^2 - 2^2} = \sqrt{96} = 4\sqrt{6}, and the two internal ones \displaystyle \sqrt{10^2 - 8^2} = \sqrt{36} = 6.
Lengths of the common tangents
Two circles have radii 9 and 4, and their centers are 15 apart. Find the length of each kind of common tangent.
SolutionHere R + r = 13 and d = 15 > 13, so the circles are apart and have all four common tangents. For an external tangent, the leg of the right triangle in Figure 4 is R - r = 5:
For an internal tangent the leg is R + r = 13:
The internal tangents are the shorter pair. They always are when both kinds exist, because the hypotenuse d is the same and the leg R + r is longer than R - r.
The triangle of contact points
In \triangle ABC, m\angle A = 70^\circ, m\angle B = 60^\circ and m\angle C = 50^\circ. Its incircle, with center I, touches \overline{BC} at D, \overline{CA} at E and \overline{AB} at F, as in Figure 6. Find the angles of \triangle DEF.
\overline{AE} and \overline{AF} are the two tangent segments from A, so in the quadrilateral AFIE the angles at E and F are right angles, and the central angle at I is the supplement of the angle at A:
\angle EDF is an inscribed angle of the incircle opening onto the arc \overset{\frown}{EF} that does not contain D, and that arc is cut off by the central angle \angle EIF. By the inscribed angle theorem it is half the central angle:
The same argument at B and at C, whose tangent segments end at D and F, and at D and E, gives the other two:
Check. 55^\circ + 60^\circ + 65^\circ = 180^\circ. In general the angle of \triangle DEF at D is 90^\circ - \tfrac{1}{2}m\angle A, so every angle of the triangle of contact points is acute, whatever the shape of \triangle ABC.
A belt around two wheels
Two wheels with radii 10 cm and 2 cm have centers 16 cm apart. A tight belt runs round the outside of both, as in Figure 7. Find the length of the belt.
The two straight runs of the belt are the external common tangents. Here R - r = 8, so each has length
To see how much of each wheel the belt covers, use the right triangle of Figure 4 again. Its hypotenuse is O_1O_2 = 16 and its leg along the radius O_1A is R - r = 8, half the hypotenuse. A right triangle with one leg half its hypotenuse has a 30^\circ angle opposite that leg (special right triangles), so the angle at O_1 is 60^\circ: the radius O_1A makes 60^\circ with the line of centers. The lower tangent is the mirror image, so the belt leaves bare the 120^\circ arc of the large wheel that faces the small one, and wraps the other 240^\circ.
On the small wheel the radius O_2B is parallel to O_1A and points the same way, so it makes 180^\circ - 60^\circ = 120^\circ with the ray from O_2 back toward O_1. The bare arc facing the large wheel is 2 \cdot 120^\circ = 240^\circ, and the belt wraps the remaining 120^\circ.
The two arcs, by arc length, are
and the whole belt is
Boundaries
The point must be outside. From a point on the circle there is exactly one tangent, the line perpendicular to the radius there. From a point inside there is none, since every line through an inside point crosses the circle twice. Equal tangent segments and the circumscribed angle need P outside the circle.
Tangent segments end at the points of contact. PA = PB is about the segments from P to where the lines touch. Two segments from P along the same lines that stop somewhere else need not be equal.
Not every polygon has an inscribed circle. Every triangle has one. A quadrilateral needs AB + CD = BC + DA; for a convex quadrilateral that condition is also enough, though that direction takes a longer proof than the one here. A rectangle that is not a square fails it, so no circle touches all four of its sides. For a polygon in general, the bisectors of all its angles must meet at one point, and that point is the center.
The area formula needs every side touched. K = \tfrac{1}{2} r p holds because every triangle from the center has height r. If some side does not touch the circle, its triangle is taller and the formula undercounts.
The counts are for two different circles. A circle and itself share every tangent. With equal radii R - r = 0, so the cases “touching from inside” and “one inside the other” cannot happen, and two different circles of the same size always have at least the two external tangents, parallel to the line of centers.
The length formulas measure between points of contact, not from a point of contact to where two tangents cross, or to where a tangent crosses the line of centers.
All of this is in a plane. In space, a point outside a sphere has infinitely many tangent lines to it, forming a cone, all with the same tangent length.
Where this goes next
The circumscribed angle is one member of a family. The angle made by two tangents, by a tangent and a secant, or by two secants meeting outside a circle is half the difference of the two arcs they cut off, which is angles made by chords, secants and tangents. For two tangents the arcs are \overset{\frown}{AB} and the rest of the circle, and half their difference is \tfrac{1}{2}\big((360^\circ - a) - a\big), which is 180^\circ - a, the result here.
Tangent lengths also join the products of segment lengths in circles: the square of the tangent length from P equals the product of the two distances from P to the circle along any secant through P.
The converse of the first result is how a tangent is drawn with compass and straightedge: to draw the tangents from P, find the points A of the circle where \angle OAP is a right angle, which the circle on \overline{OP} as a diameter supplies. Drawings of this kind are collected in constructions with circles.
Circles through a polygon’s corners and circles touching its sides make a matching pair. A quadrilateral with a circle through its four corners has opposite angles adding to 180^\circ (cyclic quadrilaterals); a quadrilateral with a circle touching its four sides has opposite sides adding to the same total. A regular polygon has both circles, and the radius of the inner one is its The apothem of a regular polygon.
Exercises
Tangent and radius
- A line is tangent at T to a circle with center O and radius 11. A point P on the line is 61 from O. Find PT.
- The tangent segment from a point P to a circle has length 40, and P is 41 from the center. Find the radius.
- A circle has center O and radius 8. A tangent segment \overline{PT}, touching at T, also has length 8. Find OP and m\angle TPO.
Two tangents from one point
In Exercises 4 to 8, two tangents from P touch a circle with center O at A and B.
- m\angle AOB = 100^\circ. Find m\angle APB.
- m\angle APB = 38^\circ. Find m\angle AOB and the measures of the minor and major arcs AB.
- PA = 2x + 7 and PB = 4x - 9. Find x and PA.
- PA = x^2 - 5 and PB = 4x. Find x and PA.
- The radius is 5 and m\angle APB = 60^\circ. Find OP and PA.
Polygons around a circle
- Quadrilateral ABCD is circumscribed about a circle, with AB = 9, BC = 12 and CD = 15. Find DA and the perimeter.
- Quadrilateral ABCD is circumscribed about a circle, with AB = 2x + 1, BC = 3x, CD = 4x - 1 and DA = x + 6. Find x and the four sides.
- A square is circumscribed about a circle of radius 7. Find the side of the square, its area, and the area inside the square but outside the circle.
- A polygon circumscribed about a circle of radius 4 has perimeter 60. Find its area.
- A regular hexagon is circumscribed about a circle of radius 3. Find its side and its area.
- A triangle with sides AB = 11, BC = 20 and CA = 13 is circumscribed about a circle, and its area is 66. Find the tangent pieces at A, B and C, and the radius of the circle.
Common tangents
- Two circles have radii 6 and 2. How many common tangents do they have when their centers are (a) 10, (b) 8, (c) 5, (d) 4, (e) 3 apart?
- Two circles have radii 10 and 4, and their centers are 10 apart. How many common tangents are there, of which kind, and how long are they?
- Two circles have radii 3 and 2, and their centers are 13 apart. Find the lengths of the external and the internal common tangents.
- Two circles of radius 5 have centers 26 apart. Find the lengths of the external and the internal common tangents.
- An external common tangent of two circles with radii 11 and 4 has length 24. How far apart are the centers?
Reasoning
- Two tangents from P touch a circle with center O at A and B. Show that m\angle PAB = \tfrac{1}{2}m\angle AOB.
- Explain why no circle can touch all four sides of a 6 by 10 rectangle.
- Show that a parallelogram circumscribed about a circle is a rhombus.
- Two circles with centers O_1 and O_2 touch at a single point T and share the tangent line there. Explain why O_1, T and O_2 lie on one line.
- One circle lies entirely inside another without touching it. Explain why the two have no common tangent.
- Show that the tangent length from a point P to a circle with center O and radius r is \displaystyle \sqrt{OP^2 - r^2}, and use it to give a second reason why the two tangent segments from P are equal.
Answers
- PT = 60. OT \perp PT, so PT^2 = 61^2 - 11^2 = 3721 - 121 = 3600.
- r = 9. The radius to the point of contact is perpendicular to the tangent, so r^2 = 41^2 - 40^2 = 1681 - 1600 = 81.
- \displaystyle OP = 8\sqrt{2} and m\angle TPO = 45^\circ. The angle at T is a right angle and both legs are 8, so \triangle OTP is an isosceles right triangle: OP^2 = 64 + 64 = 128, and its two acute angles are equal.
- m\angle APB = 80^\circ. The circumscribed angle is the supplement of the central angle: 180^\circ - 100^\circ.
- m\angle AOB = 142^\circ; the minor arc is 142^\circ and the major arc 218^\circ. The central angle is 180^\circ - 38^\circ, the minor arc has the same measure, and the major arc is 360^\circ - 142^\circ.
- x = 8 and PA = 23. The tangent segments are equal: 2x + 7 = 4x - 9, so 16 = 2x. Check: 4(8) - 9 = 23.
- x = 5 and PA = 20. Setting x^2 - 5 = 4x gives x^2 - 4x - 5 = 0, so (x - 5)(x + 1) = 0. The root x = -1 would make PB = -4, which is not a length, so only x = 5 works, and then PA = PB = 20.
- OP = 10 and \displaystyle PA = 5\sqrt{3}. OP bisects \angle APB, so m\angle APO = 30^\circ. In the right triangle OAP the leg OA = 5 is opposite the 30^\circ angle, so it is half the hypotenuse and OP = 10. Then PA^2 = 10^2 - 5^2 = 75.
- DA = 12 and the perimeter is 48. Opposite sides have equal sums, so DA = AB + CD - BC = 9 + 15 - 12, and the perimeter is 9 + 12 + 15 + 12.
- x = 3; the sides are AB = 7, BC = 9, CD = 11 and DA = 9. Opposite sides have equal sums: (2x + 1) + (4x - 1) = 3x + (x + 6), so 6x = 4x + 6. Check: 7 + 11 = 18 = 9 + 9.
- Side 14, area 196, and the region between has area 196 - 49\pi \approx 42.1. Two opposite sides are parallel tangents, and the radii to their points of contact are both perpendicular to them, so together they make a diameter running straight across: the side is 2 \cdot 7. The circle has area \pi \cdot 7^2 = 49\pi.
- 120. K = \tfrac{1}{2} r p = \tfrac{1}{2}(4)(60).
- Side \displaystyle 2\sqrt{3}, area \displaystyle 18\sqrt{3}. As in Example 4, the six triangles from the center are equilateral with height 3, so \displaystyle \frac{\sqrt{3}}{2}s = 3 and \displaystyle s = \frac{6}{\sqrt{3}} = 2\sqrt{3}. Then \displaystyle K = \tfrac{1}{2}(3)(6 \cdot 2\sqrt{3}) = 18\sqrt{3}.
- The pieces are 2 at A, 9 at B and 11 at C, and r = 3. Half the perimeter is s = 22, and the piece at each corner is s minus the opposite side: 22 - 20, 22 - 13 and 22 - 11. Check: 2 + 9 = 11, 9 + 11 = 20, 11 + 2 = 13. The area formula K = \tfrac{1}{2} r p is K = r s, so the radius is K / s = 66 / 22.
- (a) 4, (b) 3, (c) 2, (d) 1, (e) 0. Compare d with R + r = 8 and R - r = 4: 10 > 8, 8 = 8, 4 < 5 < 8, 4 = 4 and 3 < 4.
- Two, both external, each of length 8. Here R - r = 6 < 10 < 14 = R + r, so the circles cross at two points and no line passes between them. \displaystyle L_{\text{ext}} = \sqrt{10^2 - 6^2} = 8.
- External \displaystyle 2\sqrt{42} \approx 12.96, internal 12. \displaystyle \sqrt{13^2 - 1^2} = \sqrt{168} = 2\sqrt{42} and \displaystyle \sqrt{13^2 - 5^2} = \sqrt{144} = 12.
- External 26, internal 24. With equal radii R - r = 0, so L_{\text{ext}} = 26, the distance between the centers: the two radii and the tangent make a rectangle with the line of centers. \displaystyle L_{\text{int}} = \sqrt{26^2 - 10^2} = \sqrt{576} = 24.
- d = 25. The right triangle has legs 24 and 11 - 4 = 7 and hypotenuse d, so d^2 = 576 + 49 = 625.
- \triangle OAB is isosceles because OA = OB, so m\angle OAB = \tfrac{1}{2}(180^\circ - m\angle AOB), which is 90^\circ - \tfrac{1}{2}m\angle AOB. The radius OA is perpendicular to the tangent, and the chord \overline{AB} lies inside the right angle \angle OAP, so m\angle PAB = 90^\circ - m\angle OAB, which is \tfrac{1}{2}m\angle AOB.
- If a circle touched all four sides, opposite sides would have equal sums, but 6 + 6 = 12 and 10 + 10 = 20. Seen directly: a circle touching both long sides has diameter 6, the distance between them, while a circle touching both short sides needs diameter 10.
- In a parallelogram ABCD, AB = CD and BC = DA (properties of a parallelogram). Being circumscribed gives AB + CD = BC + DA, so 2AB = 2BC and AB = BC. All four sides are then equal, and the parallelogram is a rhombus.
- Each radius to the point of contact is perpendicular to the tangent line: O_1T \perp \ell and O_2T \perp \ell, both at T. Through T there is only one line perpendicular to \ell, so O_1 and O_2 both lie on it, and so does T.
- Let the inner circle have center O_2 and radius r, and the outer one center O_1 and radius R, with d + r < R. A line tangent to the inner circle passes through a point X of it, and by the triangle inequality O_1X \le O_1O_2 + O_2X, which is d + r < R. So X is inside the outer circle, and a line through a point inside a circle crosses it twice. The line is not tangent to the outer circle.
- If \overline{PA} is tangent at A, then OA \perp PA, so \triangle OAP is right-angled at A with hypotenuse OP and leg OA = r, and PA^2 = OP^2 - r^2. The other tangent segment \overline{PB} gives a right triangle with the same hypotenuse OP and the same leg r, so PB^2 = OP^2 - r^2 as well. Both lengths are positive with the same square, so PA = PB.