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Algebra

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8 Jan 2009 09:43:42 IST

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A nice result

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Prove that every circle passing through the points and intersects the circle at right angles

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Ankit Rana

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8 Jan 2009 13:27:21 IST

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no they do not since there is one such circle that just touches the circle |z| = 1 externally

and that circle is the one which is draw with the points z₀ and 1/z₀(bar) as the end points of diameters

1/z₀(bar) can be written as z₀ / |z₀|²which lies on the unit circle

Circles that touch each other externally are not orthogonal to each other

plz. correct if wrong

Hari Shankar

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8 Jan 2009 13:33:07 IST

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Why must z₀/|z₀|² lie on the unit circle?

But yes, the only exception is when |z₀| = 1

abhishek sinha

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8 Jan 2009 13:41:43 IST

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see the latexified version below :

abhishek sinha

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8 Jan 2009 14:01:16 IST

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information missing : |z0|!=1

solution : let the eqn of any circle passing through the points given by :

$|z|^2+a{\bar{z}}+{\bar{a}z+b=0$

which has origin at -a and radius = |a|^2 -b

Now this eqn satisfies z=z0 , and z= $\frac{1}{\bar{z_0}}$

first condition gives

$|z_{0}|^2+a{\bar{z_{0}}+\bar{a}z_{0}+b=0$ .............(1)

the second condition gives

$b|z_{0}|^2+a{\bar{z_{0}}+\bar{a}z_{0}+1=0$ ................(2)

comparing (1) and (2) and utilising the condition that | z_0 | !=1,

we get , b=1

Now square of distance between the origin of the above circle and {|z|=1} is = |a|^2 = |a|^2 -1 +1

sum of the square of their individual radii .

Hence they cut orthogonally . ( Proved)

Hari Shankar

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8 Jan 2009 14:01:44 IST

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actually since the equation of circles etc. in complex coordinates is not in jee syllabus, this isnt the solution i am looking for.

abhishek sinha

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8 Jan 2009 14:07:38 IST

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It is a simple matter to prove the equation of the circle in complex co-ordinates ..

Suppose we have to find the eqn of circle with centre at c and radius =r

then we know that its eqn is

|z-c|^2=r^2

Now utilise the relation $|z-c|^2=(z-c) (\bar{z}-\bar{c})$

which when expanded reduces to the above form...

btw . I think that this simple thing is well within the JEE syllabus !

Ankit Rana

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8 Jan 2009 14:18:24 IST

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Hari Shankar

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8 Jan 2009 14:19:37 IST

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@feynmann: well its nice of you to give the equation. nice solution too. Now, if any of the others could come up with a less synthetic solution. This is not a challenge. Its a nice instance of geometry in complex numbers

Hari Shankar

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8 Jan 2009 14:23:50 IST

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good work ankit. now, you have done it algebraically. So you are in the best position to compare it with a geometric solution. You have brought out one very useful fact, that and origin, z₀ and $\frac{1}{\bar{z_0}}$ are collinear.

Do give it a try!

PS: I could do it without having to put pen to paper. That should tell you how easy the solution by geomtetry is.

Ankit Rana

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8 Jan 2009 14:24:49 IST

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thanx

sebin

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8 Jan 2009 18:57:50 IST

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clearly if r >1 , 1/r<1 and origin Z and 1/z(bar) are collinear

two circles orthogonal if ( and are their radi and their centres at a distance )

clearly r1=

and r2=1

and d=

so we have to prove that

Lhs = =Rhs

Hari Shankar

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8 Jan 2009 19:13:53 IST

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Yup done. that's all u need. No fancy circle formulae required as you can see. Also you really get a feel of what's going on. Good work sebin

I think this problem really demonstrates how powerful geometric interpretation is. It totally dominates algebraic method in this case. So, more often than not, if you interpret the problem in a geometric context, you will reap rich awards.

Hari Shankar

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8 Jan 2009 19:20:02 IST

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Also, a little bit of extra reading helps in unexpected ways.

If O is the centre of a circle with radius r and P is some point, then the point P' lying on OP (extended if required) such that OP . OP' = r² is called the inversion point with respect to this circle. Several interesting geometric properties flow out of this. And you can see that with |z| = 1, $\frac{1}{\bar{z_0}}$ is the inversion of z₀

I wonder if I could have cottoned on easily to the fact that $\frac{1}{\bar{z_0}}$ is collinear with z₀ and origin,if I hadnt been aware of this concept.

sebin

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8 Jan 2009 19:40:15 IST

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thank you sir

Anant Kumar

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8 Jan 2009 21:36:43 IST

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There is only a slight glitch in your proof sebin. The question says "every circle passing through $z_0$ and $\dfrac{1}{\overline{z}_0}$ " and not the only circle having these two complex numbers as ends of diameter.

Having said that, however, the same method method can be adapted to prove the general case when the segment joining $A(z_0)$ and $B(1/\overline{z}_0)$ is a chord. Since $\dfrac{1}{\overline{z}_0}=\dfrac{z_0}{|z_0|^2}$ , it follows that $\arg(z_0)=\arg \left(\dfrac{1}{\overline{z}_0}\right)$ . Therefore, the origin $O$ , $A$ and $B$ lie on a straight line.

Further, if $z_0$ lies inside the unit circle, then $\dfrac{1}{\overline{z}_0}$ will necessarily be outside the unit circle, and vice-versa. So I prove for the case when $|z_0|<1$ .

The figure shows a circle centered at $C$ passing through $A$ and $B$ . Let the center $C$ represent the complex number $a$ . The two circles intersect at two points, one of which is $P$ . The two circles will be orthogonal iff

$OC^2 = OP^2 + PC^2=1+AC^2$

$\Leftrightarrow \ |a|^2 = 1+|a-z_0|^2=1+|a|^2+|z_0|^2 -2\mathrm{Re}(a\overline{z}_0)$

$\Leftrightarrow \ |z_0|^2 + 1 -2\mathrm{Re}(a\overline{z}_0)=0\qquad(1)$

Since the center $C$ of the circle must lie on the perpendicular bisector of $AB$ , we must have

$AC=BC\quad\Rightarrow AC^2=BC^2$

$\Rightarrow \ |a-z_0|^2 = \left|a-\dfrac{1}{\overline{z}_0}\right|^2$

$\Rightarrow \ |a|^2 + |z_0|^2 -2\mathrm{Re}(a\overline{z}_0) = |a|^2 + \dfrac{1}{|\overline{z}_0|^2}-2\mathrm{Re}\left(\dfrac{a}{z_0}\right)$

$\Rightarrow \ |z_0|^2 -2\mathrm{Re}(a\overline{z}_0) = \dfrac{1}{|z_0|^2}-2\mathrm{Re}\left(\dfrac{a\overline{z}_0}{|z_0|^2}\right)$

$\Rightarrow \ |z_0|^2 -2\mathrm{Re}(a\overline{z}_0) = \dfrac{1}{|z_0|^2}-2\dfrac{\mathrm{Re}(a\overline{z}_0)}{|z_0|^2}$

$\Rightarrow\ |z_0|^4 -2\mathrm{Re}(a\overline{z}_0) |z_0|^2 + 2\mathrm{Re}(a\overline{z}_0) -1 =0$

$\Rightarrow \ (|z_0|^2 -1)(|z_0|^2 + 1 -2\mathrm{Re}(a\overline{z}_0)) = 0$

But since $|z_0|^2-1\neq 0$ , we get $|z_0|^2 + 1 -2\mathrm{Re}(a\overline{z}_0)=0$ which is precisely we wanted.

Hari Shankar

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8 Jan 2009 22:10:07 IST

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I did see that. I was so giddy with relief that someone did solve it that i didnt have the heart to point that out. But that's only a short extension. Again only Pythagoras needed. What we must remember is that centre of the circle will lie on the perpendicular to the chord joining z₀ and $\frac{1}{\bar{z_0}}$ . Let's say that its at a distance h along the perp.Then , the square of the distance between the centres now becomes $\frac{1}{4} \left(r+\frac{1}{r} \right)^2 + h^2$

and the square of the radius of the circle becomes $\frac{1}{4} \left(r-\frac{1}{r} \right)^2 + h^2$

Hence we still have the equality $\frac{1}{4} \left(r-\frac{1}{r} \right)^2 + h^2+1 = \frac{1}{4} \left(r+\frac{1}{r} \right)^2+h^2$ and we are done

Actually it was this part of the result that made me really like it. Its true not just for the circle that has those points as the diameter, but any circle passing thru them. Also every circle passing through those points does intersect |z| = 1. All this made me really go ga ga!

Hari Shankar

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8 Jan 2009 22:16:31 IST

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have to add, here r = |z₀|

Also, thank you kaymant sir for that lovely diagram. All these arguments become simple to follow with the diagram as an aid

Dipanjan

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10 Jan 2009 12:44:14 IST

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Re:A nice result

Hari Shankar

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10 Jan 2009 13:32:02 IST

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thats truly fantastic! Well thats Geometry Methods 2 Algebraic Methods 0

great contribution jishnu!!!!!

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