Two Circles and Families of Circles

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Draw two circles on a page. How many straight lines can touch both of them at once? Is it four, three, or maybe none at all? The answer changes as one circle slides towards the other, and just one distance decides it. Then we find the line through their meeting points, and whole families of circles. These ideas come up often in JEE, so let's begin.

0:28Two circles: one distance decides

So, C sub one and C sub two are the centres; the small one and two below the letters are subscripts, just labels. r sub one and r sub two are the radii, and d, written C one C two, is the distance between the centres. Okay? Let me ask you this: when are the two circles completely apart? Compare d with the radii. Yes, exactly, when d is greater than r one plus r two; the open side of the sign faces the bigger quantity. Then the circles are separate, with four common tangents, four lines that touch both circles. With me? Now slide them closer, until d equals r one plus r two. They touch from outside, and the count drops to three. Right? Closer still, and they cut at two points. The vertical bars around r one minus r two mean mod, the modulus, so the difference is positive whichever radius is bigger. With d between these two values, the circles intersect, and two common tangents are left. Clear? When d equals the modulus of r one minus r two, the smaller circle touches the bigger one from inside, and there is just one common tangent. Got it? And when d is less than that modulus, one circle lies completely inside the other, so there are zero common tangents. Notice the countdown as the circles come closer: four, three, two, one, zero. So the answer to our opening puzzle depends on how d compares with the radii. Okay?

1:56Common chord: subtract the equations

Now write both circles in the general form. S one is x squared plus y squared, plus two g one x, plus two f one y, plus c one; the sign with three lines means identically equal. S two is the same with g two, f two and c two. Here, S one and S two are just names for the two circles. With me? Now subtract S two from S one. What happens to the x squared and y squared terms? Yes, they cancel! What is left is a straight line, S one minus S two equals zero. Any point on both circles makes S one and S two zero, so it lies on this line too. So when the circles cut, this line is the common chord, also called the radical axis. Right? One warning before you subtract: x squared and y squared need coefficient one in both equations, as on screen, so divide first if they do not. Clear?

2:49Orthogonal circles: cutting at right angles

Next, orthogonal circles. They cut at right angles, which means that at a meeting point, the two tangents are perpendicular. Okay? Now, a radius is always perpendicular to the tangent at its end. So what is the angle between the two radii at that point? Correct, it is a right angle too! So the two centres and the meeting point make a right triangle, with sides r one and r two, and hypotenuse d. Pythagoras gives d squared equals r one squared plus r two squared. Got it? From the general form, the centres are minus g one comma minus f one, and minus g two comma minus f two, and each radius squared is g squared plus f squared minus c. With me? Put these in and expand; almost everything cancels, leaving two g one g two, plus two f one f two, equals c one plus c two. Careful here, the right side is a sum, not c one times c two. Okay?

3:44Families of circles

Now, families of circles. Take a circle, S equals zero, and a line, L equals zero, that cuts it, and form S plus lambda times L equals zero. Lambda, the Greek letter, is any number we choose. Each meeting point makes S and L both zero, so it satisfies this for every lambda. And x squared plus y squared stay untouched, so each one is a circle. One equation, a whole family. Clear? The same idea works for two circles: S one plus lambda times S two equals zero. Then one more condition, like a point on the circle, fixes lambda. Right? But there is one condition: lambda must not be minus one. Can you see why? Put lambda equals minus one and see. Yes, exactly! You get S one minus S two equals zero, the common chord, a line, not a circle. The equals sign with a slash through it means not equal to. Got it?

4:38Example 1: common tangents of circles touching externally

Now one from the course. How many common tangents do x squared plus y squared equals nine, and, x minus five, whole squared, plus y squared equals four, have? Pause the video and try it first. Done? Okay, let's check. The first circle has centre zero comma zero and radius three. The second has centre five comma zero and radius two. Okay? Both centres lie on the x axis, so d is five. And r one plus r two is three plus two, also five. Right? So the circles touch. Then there is only one common tangent, the one at the point of contact, isn't it? No, no, careful, that is exactly the trap! They touch externally, so the two outer tangents are still there, along with the one at the point of contact. So the answer is three common tangents. Only touching internally gives one. Clear?

5:32Example 2: circle through x2 + y2 = 4, x + y = 1 and (1, 1)

Now a slightly harder one. Find the circle through the intersections of x squared plus y squared equals four and the line x plus y equals one, which also passes through one comma one. Pause the video and try it first. Here is a hint: use a family of circles. Done? Okay, let's check. Finding the meeting points first is the long way. We have a circle and a line, so which family do we use? Yes, exactly, the circle plus lambda times the line! With everything on one side, that is x squared plus y squared minus four, plus lambda times, x plus y minus one, equals zero. With me? It passes through one comma one, so put x equal to one and y equal to one. So, one plus one minus four, plus lambda times, one plus one minus one, equals zero. Okay? That is minus two plus lambda equals zero, so lambda equals two. Nice and quick, isn't it? Put lambda equals two back: x squared plus y squared minus four, plus two times, x plus y minus one, equals zero. Clear? Open the bracket: minus four minus two gives minus six. So the circle is x squared plus y squared, plus two x, plus two y, minus six, equals zero. Quick check: put in one comma one, and one plus one plus two plus two, minus six, gives zero. It works perfectly.

7:00Quick revision: tangent count, common chord, families

So, quick revision. Compare d with r one plus r two and with the modulus of r one minus r two, and the common tangents count down four, three, two, one, zero. Subtract the equations for the common chord, and for orthogonal circles, two g one g two plus two f one f two equals c one plus c two. For circles through meeting points, use S plus lambda L, or S one plus lambda S two, with lambda not equal to minus one. Now try the practice problems on this page, and all the best!