Why the two from an outside point are always equally long

0:00We finished last time with a measurement and no argument. From a point outside a circle there are exactly two tangents. Measure the piece of each one from the point to where it touches, and the two come out the same length. Move the point. Same again. Change the radius, move the circle, try it wherever you like. The two are always equal. That is a pattern, and a pattern is not a proof. Every circle you have time to draw is still a finite number of circles.

0:35So this time we are going to show it must hold, for every circle and every outside point at once, and we are going to do it with something we already own. The whole proof is one theorem, used twice. Here is the picture. A circle with centre O, a point P outside it, and the two tangents from P touching at Q and at R. As it stands there is nothing to work with. Two tangent segments and a circle.

1:07So we add three segments that were not there: from P to the centre, from the centre out to Q, and from the centre out to R. Notice what those three are. Two of them are radii. The third joins the two points the whole question is about. And look at what the picture has become. Two triangles, O Q P and O R P, sharing the segment O P.

1:33That sharing is the part that never changes. Whatever else moves, both triangles hang off the same segment. Now the theorem we already have, the one that says a tangent meets the radius at the point of contact at a right angle. P Q is a tangent and O Q is the radius to its point of contact. So the angle at Q is a right angle. P R is a tangent and O R is the radius to its point of contact. So the angle at R is a right angle too.

2:11One theorem, applied at two places. That is the only thing being imported into this proof, and it is worth saying plainly, because everything that follows is arithmetic and bookkeeping on top of it. Both triangles are right triangles now, and in both of them the right angle sits at the touching point. Which means the shared segment O P is the hypotenuse of each. Not a side that happens to be shared. The hypotenuse of both.

2:45So take stock of what the two triangles agree on. The hypotenuse: it is the same segment, O P, in both. Not equal by measurement. The same segment. One leg: O Q and O R are both radii of the same circle, so they are equal. And the right angle sits in the same place in each, at the touching point, facing that hypotenuse. Hypotenuse and one side, in two right triangles. That is exactly the congruence rule called R H S, and it says the two triangles are congruent.

3:25Once they are congruent, every matching part agrees, so the remaining legs agree. P Q equals P R. That is the theorem, and it took four lines. It went by quickly, so let us slow down at the one step that is doing real work. We matched two sides and an angle. But the angle is not between the two sides we matched. It is at the far end, facing the hypotenuse.

3:55In general that is not enough. Two sides and an angle that is not between them can leave you with two different triangles. Here is why. Fix one corner and one side out of it, and set the angle you want. That fixes a ray for the third corner to sit on. Now the third corner has to be a stated distance from the far end of that side. So it has to be where the ray meets a circle of that radius, and a ray can meet a circle twice.

4:27Two places for the third corner. Two triangles. Same two sides, same angle, different shapes. Unless the angle is a right angle. Then the start of the ray is already the closest point of the ray to that far end, so the ray can only reach any given distance once. One place, one triangle. That is the whole of what R H S is given, and why the right angle is not decoration here.

4:59There is a second route to the same conclusion, and it shows something the first one hides. In the right triangle at Q, the hypotenuse is O P and the legs are the radius and the tangent length. So the tangent length squared is O P squared less the radius squared. In the right triangle at R, the hypotenuse is O P and the legs are the radius and the tangent length. So the tangent length squared is O P squared less the radius squared.

5:34Those are not two similar-looking calculations. They are the same expression, and the two numbers going into it, the distance to the centre and the radius, are the same two numbers both times. So the equality is not a coincidence between two triangles. It is one calculation carried out twice on the same inputs. One thing that expression does not fix, by the way, is which of the two legs is the shorter one. The tangent beats the radius only when the point stands further out than the radius times root two.

6:14Closer in than that and the radius is the longer leg. It happens that the examples people set are almost always well outside, which quietly teaches the wrong habit. The congruence gives more than we asked it for, and it costs nothing extra. Congruent triangles match on every part, not just the side we wanted. So look at the angles at P. The angle between P Q and P O equals the angle between P R and P O.

6:49In other words, the line from the outside point to the centre cuts the angle between the two tangents exactly in half. The centre sits on the bisector of that angle. That is not a second theorem to learn. It is the same congruence read on a different pair of corresponding parts, which is worth knowing because most later questions reach for this half rather than for the equal lengths.

7:18Two results, one proof, and the second one was free. Now what the equality is actually for. Here is the first of four uses, and it is the prettiest. Two circles with the same centre, one inside the other. Draw a chord of the big one that just grazes the small one, touching it at a point P. Claim: P is the exact midpoint of that chord. The trick is that the same segment is playing two roles at once. Against the small circle it is a tangent. Against the big circle it is a chord.

7:57As a tangent to the small circle, the radius to P meets it at a right angle. That is the theorem again. And a segment from the centre meeting a chord at a right angle cuts that chord in half. So P is the midpoint. Each role contributed one step. With radii five and three, each half comes out as four, because twenty five take away nine is sixteen, so the chord is eight long.

8:28Second use. Keep P outside, tangents touching at Q and at R, and join Q to R. Call the angle at P between the two tangents theta. The claim is that the angle at Q, between that joining segment and the radius O Q, is exactly half of it. Here is the working. P Q and P R are equal, which we now know, so the triangle on P, Q and R is isosceles.

8:59Its two base angles are therefore equal, and the three angles total a hundred and eighty, so each base angle is ninety less half of theta. Now the angle at Q between the tangent and the radius is a right angle, ninety. The angle we want is what is left when the base angle is taken out of that right angle. Ninety, less ninety less half of theta. The ninetys cancel and half of theta is what survives. Notice the shape of that: the equal lengths were the hypothesis, not the conclusion. This is the first place the theorem earns its keep.

9:42Third use, and this one turns three separate questions into one. Take the four-sided figure O, Q, P, R. Its corners are the centre, one touching point, the outside point, and the other touching point. Two of its four angles are already known. The angle at Q is a right angle and the angle at R is a right angle, both for the same reason as before. The four angles of a four-sided figure total three hundred and sixty. Take away the two right angles and a hundred and eighty is left.

10:21So the angle at the outside point and the angle at the centre always add to a hundred and eighty. They are supplementary, whatever the circle and wherever the point. A hundred and ten at the centre leaves seventy at the point. Read it the other way and eighty at the point leaves a hundred at the centre. And the commonest slip in this whole topic is to make those two add to ninety instead. They do not. Ninety is what the two HALVES make, which is a different configuration entirely.

10:58Fourth use, and now some arithmetic. A circle of radius five carries a chord eight long. The tangents at the two ends of that chord meet at a point. How far is that point from each end? Route one. The two tangent lengths are equal, so the triangle on the meeting point and the two ends is isosceles, and the line to the centre bisects its apex angle. That line therefore meets the chord at a right angle and halves it, so each half is four.

11:32The reach from the centre to the chord is then three, because five squared less four squared is nine. Two of the right triangles in that picture share an angle as well as a right angle, so they are similar, and the sides in one are in the same ratio as the sides in the other. The tangent is to five as four is to three. Which gives twenty thirds. About six and two thirds.

12:00Route two, with no similar triangles at all. Call the tangent length x and the piece of the line from the meeting point to the chord y. One right triangle gives x squared as y squared plus sixteen. The other gives x squared plus twenty five as y plus three, all squared. Subtract the first from the second and everything squared disappears. Six y is thirty two, so y is sixteen thirds, and x comes out at twenty thirds again. Two routes, no shared working, one answer.

12:39The last use is the one that looks like a different subject. Draw a four-sided figure round a circle, so that all four sides touch it. Each corner is an outside point with two tangents, and the two tangent lengths from that corner are equal. So colour them. Each corner gets its own colour, and each side is made of two coloured pieces, one from each end. Now add up one pair of opposite sides. You get all four colours, once each.

13:13Add up the other pair of opposite sides. You get all four colours, once each. Same four pieces both times, so the two totals must agree. Opposite sides of a four-sided figure drawn round a circle add to the same total, and the argument was really just sorting. The hypothesis is doing genuine work, by the way. A rectangle that is not a square has no circle touching all four of its sides, and its opposite sides do not add to the same total. Which is also why a parallelogram drawn round a circle has to be a rhombus.

13:56One more, with numbers, and then we are done. A triangle drawn round a circle of radius four. The circle touches the longest side at a point that cuts it into eight on one side and six on the other. Find the other two sides. Call the two equal pieces at the top corner x. Then the side down the left is x plus eight and the side down the right is x plus six, and the bottom is fourteen.

14:27Now the area, twice. The radius reaches every side at a right angle, so the area is the radius times half the perimeter, which is four times x plus fourteen. The area from the three sides alone is the square root of x plus fourteen, times x, times eight, times six. Set the two equal, square both sides, and one factor of x plus fourteen cancels off each. Sixteen x plus two hundred and twenty four is forty eight x, so x is seven.

15:04Which makes the sides fifteen, fourteen and thirteen. Worth a check: those three give an area of eighty four and a half perimeter of twenty one, and eighty four over twenty one is four, the radius we were given. Four questions, and the same one fact under all of them.