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Euclidean Formula for Orthogonal Circles

Date: 04/11/2001 at 19:31:19
From: Alisen
Subject: Euclidean formula for orthogonal circles

We are having a hard time figuring this out. We know that the 
equation for a circle C in the Euclidean plane with radius r and 
center (h,k) is : (x-h)^2 + (y-k)^2 = r^2 then expanded is : 
x^2-2hx+y^2-2ky+h^2+k^2=r^2. When considering the case when C has 
center at the origin and a radius 1, we need to show that the equation 
of the circle orthogonal to circle C and with center (h,k) is given 
by: x^2-2hx+y^2-2ky+1=0. 

I am not sure really where to begin. Your help is greatly appreciated! 

Date: 04/12/2001 at 15:12:08
From: Doctor Floor
Subject: Re: Euclidean formula for orthogonal circles

Hi, Alisen,

Thanks for writing.

We have the unit circle C 

    x^2 + y^2 - 1 = 0.................[1]

with center O(0,0) and a circle C' perpendicular to C with center 

Let X and Y be the points where the two circles meet. The tangents in 
X to C and C' are perpendicular, and the same is true in Y. But since 
each tangent to a circle is perpendicular to its radius, the radii OX 
and PX must be perpendicular, as well as OY and PY. But then OXP and 
OYP are right triangles. 

It is now simple to compute lengths PX and PY by the Pythagorean 
theorem. Since PO^2 = h^2 + k^2 and OX = 1 (radius in C) we find 
PX^2 = h^2 + k^2 - 1. And thus the circle C' has equation

  (x-h)^2 + (y-k)^2 = h^2 + k^2 - 1

which rewrites to the desired equation.

A different approach: 

The fact that OXP and OYP are right triangles also shows us that X and 
Y lie on a circle C" with OP as diameter.

This circle C" has center (h/2,k/2) and its equation is

  (x-h/2)^2 + (y-k/2)^2 = (h^2 + k^2)/4

and expanded this gives

  x^2 - hx + y^2 - ky = 0 ........... [2]

Now we use a trick that makes things easy. Any linear combination of 
the equations [1] and [2] of circles C and C" again gives a circle, 
and it passes through the same points of intersection X and Y. By this 
we don't have to actually compute the coordinates of X and Y.

One of the linear combinations is 2*[2]-[1]. This reads:

  x^2 - 2hx + y^2 - 2kx + 1 = 0

which is the desired equation, and also the circle through X and Y 
with center P(h,k).

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum   
Associated Topics:
College Conic Sections/Circles
High School Conic Sections/Circles

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