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Topic: Circle equation
Replies: 2   Last Post: Jul 16, 1997 10:01 AM

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Steve Monson

Posts: 224
Registered: 12/6/04
Re: Circle equation
Posted: Jul 16, 1997 10:01 AM
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Randy Haas <> wrote:
>Does any one know of the equation of a circle that goes through 2 points
>and is tangent to a vertical line. There will be 2 circles that solve the
>| *
>| *
>Thanks for any help
>Randy Haas

If we call the points P1 and P2, then we have two equations right off the

(1) (x1-h)^2 + (y1-k)^2 = r^2
(2) (x2-h)^2 + (y2-k)^2 = r^2

If we expand the squares and subtract one equation from the other, we get

(3) (x1-x2)h + (y1-y2)k = 0

The center of the circle (h,k) lies on the perpendicular bisector of the
line through P1 and P2. If we let P3 = ((x1+x2)/2,(y1+y2)/2), the
center of this line segment, the equation of the perp. bisector is

(y-y3)/(x-x3) = -(x2-x1)/(y2-y1) = -1/m

where m is the slope of the line joining P1 and P2.

Plugging (h,k) and rearranging a bit, we have

(4) h + mk = x3 + my3

We may without loss of generality assume the circle is tangent to the
y-axis. If not, just substitute h = h+v at the end, if the vertical line
was x=v. So,

h = r

Plugging that into (4) and (3), we get a quadratic in r:

(5) 1/m^2 r^2 - 2(y1/m + x1)r + (x1^2 + y1^2) = 0

That will give us 2 values for r, as we expected. From that it is
possible (note: I did not say easy) to compute h and k.

I'll let you do that, and maybe simplify things if I've let things
get out of hand. I may have missed some simplifying assumption.

Steve Monson
Q: Why did Arnold Schwartzenegger and Maria Shriver get married?
A: They're trying to breed a bullet-proof Kennedy.

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