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can a circle have an imaginary radius?

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Solar^

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Jan 28, 2006, 5:11:45 PM1/28/06
to
In pre-calc were doing equations of circles, anyway have come up with
the equation for a circle that involves taking the square root of a
negative number to determine the radius. Can a circle have an
imaginary radius? The entire equation is as follows in case someone is
interested:
x^2 + y^2 - 6x + 4y + 13 = 0
(x-3)^2 + (y+2)^2 = -2
so the center is C(3,-2) radius = square root of -2 or i * square
root of 2
Regards,
Solar^

N. Silver

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Jan 28, 2006, 5:30:22 PM1/28/06
to
Solar^ wrote:

> Can a circle have an imaginary radius?

Usually a radius is a distance (from the center),
and distances usually are represented by non-negative
real numbers.

One can graph the given equation from the complex
(numbers) plane to the complex plane, which would
require 4 dimensions.

> x^2 + y^2 - 6x + 4y + 13 = 0
> (x-3)^2 + (y+2)^2 = -2
> so the center is C(3,-2) radius =
> square root of -2 or i * square root of 2

If you're work were correct, then in the xy-plane
there would be no solutions. What you have derived
and concluded, formally, is nonsense. But it is also
incorrect.

x^2 + y^2 - 6x + 4y + 13 = 0

(x-3)^2+(y+2)^2 = 0,
which is satisfied by
x = 3 and y = -2.

Solar^

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Jan 28, 2006, 5:45:39 PM1/28/06
to

Thanks for the reply,
So what you are saying is that the equation DOES NOT represent a
circle? How should this be presented in homework? As no solution or ?

Stan Brown

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Jan 28, 2006, 5:56:17 PM1/28/06
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Sat, 28 Jan 2006 16:11:45 -0600 from Solar^ <nos...@nospam.com>:

y^2 There are more maths than are dreamt of in my philosophy, so I
suppose an imaginary radius is possible in somne form of advanced
math, but I've never heard of anything like that.

If you come up with an imaginary radius in your pre-calculus class, I
can say confidently that _either_ there was a mistake somewhere _or_
you're expected to identify it as "impossible".

However, the particular example you gave does not come up with an
imaginary radius:

x^2 + y^2 - 6x + 4y + 13 = 0

x^2 - 6x + 9 + y^2 + 4y + 4 = 0
(x-3)^2 + (y+2)^2 = 0

This is a "circle" of radius 0, i.e. the single point (3,-2).

--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com/

Solar^

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Jan 28, 2006, 5:57:25 PM1/28/06
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Sorry,
I see my mistake now. The equation just represents a point (3,-2)
Regards,
Solar^

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