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## #1 2006-03-15 01:35:23

RickyOswaldIOW
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### Tangent to a Circle

I need to find the equations of tangents on circles at certain points.
I have the co-ords of the point(p) and the co-ords of the center(c).  To work out the gradient of the radius from the center to the point I do:
y(c) - y(p) / x(c) - x(p)
Is this correct or should I take (p) - (c) instead?

Aloha Nui means Goodbye.

## #2 2006-03-15 01:39:51

RickyOswaldIOW
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### Re: Tangent to a Circle

x² + y² = 8 at (2, 2)
(2)² + (2)² - 8 = 0 (satisfy that the point lies on the circle)

x² + y² = 8
(x + 0)² + (y + 0)² = 8
center point is (0, 0).

0 - 2/0 - 2 = -2 / -2 = 1
the gradient of the radius is 1 and thus the gradient of the tangent is -1.

y - 2 = -1(x - 2) -->  y = x + 4.

Aloha Nui means Goodbye.

## #3 2006-03-15 04:56:43

Ikcelaks
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### Re: Tangent to a Circle

Your equation for the gradient of the radius is correct.  Note that it doesn't matter if you take p - c or c - p as long as you are consistent (sign switches in both numerator and denominator and thus cancels).

## #4 2006-03-16 08:39:05

RickyOswaldIOW
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### Re: Tangent to a Circle

Note that it doesn't matter if you take p - c or c - p

Of course

Aloha Nui means Goodbye.

## #5 2006-03-16 08:48:17

RickyOswaldIOW
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### Re: Tangent to a Circle

edit:

y - 2 = -1(x - 2) -->  y = -x + 4.

Aloha Nui means Goodbye.

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