Find the equation of the circle with diameter AB where A is (4, 7) and B is (–2, 3)

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- Dec 5th 2008, 12:32 AM #1

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- Dec 5th 2008, 01:29 AM #2

- Dec 6th 2008, 12:36 AM #3

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- Dec 6th 2008, 01:10 AM #4

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I took the liberty and wrote an explanation. I hope no one has any objections.

Before starting, I want to point out a mistake in 4. We have found the radius to be $\displaystyle \frac{\sqrt{52}}{2} = \frac{\not{2}\sqrt{13}}{\not{2}}$. Therefore, $\displaystyle r^2$ is 13 and not 13/2.

To find the equation of the circle with the given information, we need two things: the magnitude of the radius and the coordinate of the center. Recall that the equation of the circle is of the form:

$\displaystyle (x-h)^2+(y-k)^2 = r^2$

where (h, k) are the coordinates of the center and r is the radius.

1. Taking the average of the individual x and y coordinates yields the midpoint of diameter AB, which is also the center of the circle.

2. Using the distance formula:

$\displaystyle d^2 = (x_2 - x_1)^2 + (y_2-y_1)^2 $

we find the measure of the diameter AB. Dividing it by 2 yields the radius.

3 & 4. Replacing in the equation of the circle yields:

$\displaystyle (x-1)^2+(y-5)^2 = 13$