Mixed Problems: Set 2 (Circles)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. From the point four comma zero, two tangents are drawn to the circle x squared plus y squared equals eight. What is the angle between them? Many students find both slopes, then the angle between two lines, and two minutes are gone. But there is a way to get it in about five seconds, and we will reach it today. This is Set two of our mixed problems: eight circle problems, about two minutes each. Circle questions come up often in JEE Main, so let's begin!

0:36Set 2: pick the right tool first

So, first let us collect the tools; each problem in this set needs one or two of them. The length of the tangent from the point x one, y one, in round brackets, is root S one; the small one below is a subscript, and the tick shape is the square root sign. S one means: put x one and y one into the circle's equation, written with the coefficients of x squared and y squared equal to one; the small two up top means squared. Okay? Second, a line touches the circle exactly when d, its distance from the centre, equals r, the radius. No quadratic to solve. Right? Third, if a chord is at distance d from the centre, its length is two times the root of, r squared minus d squared. Clear? Fourth, from a point outside the circle, the chord joining the two points of contact is the chord of contact, and its equation is T equals zero, the same T you met in the circle chapter. With me? Now, the last tool is for two circles. If two circles cut each other, how would you get the line through their two meeting points, the common chord? Yes, exactly, just subtract one equation from the other, so S one minus S two equals zero. But careful, here S one and S two simply name the two circles' equations; no point is put in. And both must have the coefficients of x squared and y squared equal to one, before you subtract. Got it?

1:58Topper's trick: tangents at right angles

Now, back to our puzzle. The circle x squared plus y squared equals eight has its centre, C, at the origin, and its radius, r, is root eight, which is two root two. In the figure, a general sketch not drawn to scale, P is the point four comma zero, and the two orange tangents touch the circle at A and at B, okay? The dashed line joining them is the chord of contact, T equals zero. Now look at the triangle CAP in the figure. Where is its right angle? It is at A, yes, because the radius always meets the tangent at a right angle; on screen, the angle sign and the small circle, the degree sign, say that angle CAP is ninety degrees. Now let phi, the Greek letter, be half of the angle between the tangents. Then sine phi is r over C P, that is, two root two over four, which is one over root two. With me? So phi is forty five degrees, and the angle between the tangents is twice that, ninety degrees. Puzzle solved, in three lines. Clear? Now the five second way, a topper's trick. For the circle x squared plus y squared equals a squared, the director circle, x squared plus y squared equals two times a squared, is the locus of points from which the two tangents are perpendicular. Here a squared is eight, so it is x squared plus y squared equals sixteen. Four squared plus zero squared is sixteen, so P lies on it, and the angle is ninety degrees at once.

3:31Example 1: tangent length from (5, 5) using root S1

Got it? Now, Example one, an easy one from the set. Find the length of the tangent from the point five comma five to the circle x squared plus y squared, minus two x, minus four y, minus four, equals zero. Start your clock, then pause the video and try it first. Done? Okay, let's check. The coefficients of x squared and y squared are both one, so S one is ready to use. Just put in x equals five and y equals five. Okay? Twenty five plus twenty five, minus ten, minus twenty, minus four. Fifty, minus thirty four, gives sixteen. So S one is sixteen, greater than zero; the sign opening to the left means greater than. So the point is outside, and a tangent exists. Right? So the length of the tangent, the side labelled root S one in the figure, is root sixteen, which is four. One line, and done. Clear? Now check it the long way, with the figure. The centre, C, is one comma two, the radius is three, and C P comes out as five. So in the right triangle CAP, the tangent is the root of, five squared minus three squared, which is four. Same answer, but in four steps. Got it?

4:49Example 2: common chord of two circles and its length

Now a harder one, marked medium in the set. Find the common chord of the circles x squared plus y squared equals twenty five, and x squared plus y squared minus eight x equals zero, and find its length too. Pause the video and try it. Here is a hint: subtract the equations. Done? Okay, let's check. Both circles have x squared and y squared with coefficient one, good. So write each with zero on the right, and subtract: x squared plus y squared minus twenty five, minus the bracket, x squared plus y squared minus eight x, equals zero. With me? The squares cancel, and we are left with eight x minus twenty five equals zero. So the common chord is the vertical line x equals twenty five over eight. Clear? Now the length. The chord is twenty five over eight from the origin, and one of our circles has radius four. So we use two root of, sixteen minus six hundred and twenty five over sixty four, right? Oh no, careful, that mixes up the two circles! Twenty five over eight is the distance from the origin, the centre of the first circle, and that circle has radius five. Always pair a centre with its own radius. Okay? So the length is two root of, twenty five minus six hundred and twenty five over sixty four. Twenty five is one thousand six hundred over sixty four, so inside the root we get nine hundred and seventy five over sixty four. Root sixty four is eight, and two over eight is one over four, so this gives root nine hundred and seventy five, over four. And nine hundred and seventy five is twenty five times thirty nine, so the length is five root thirty nine, over four. Got it? Now a quick check with the second circle: its centre is four comma zero, and its radius is four. Its distance from the chord is four minus twenty five over eight, which gives seven over eight. And sixteen minus forty nine over sixty four is again nine hundred and seventy five over sixty four. Same length from either circle, as it must be. Right?

7:01Quick revision: tangent length, chords, director circle

So, quick revision. The tangent length is root S one; a line touches when d equals r; and a chord has length two root of, r squared minus d squared. The chord of contact is T equals zero, and the common chord is S one minus S two equals zero. And the trick: a point on the director circle gives tangents at ninety degrees; the double arrow means gives. Now start your clock, and try the rest of Set two and the practice problem on this page. All the best!