help please

**Koops** · December 5th 2008, 12:32 AM

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

**Arch_Stanton** · December 5th 2008, 01:29 AM

$S=(\frac{x_A+x_B}{2};\frac{y_A+y_B}{2})=(\frac{4-2}{2};\frac{7+3}{2})=(1,5)$
$r=\frac{|AB|}{2}=\frac{\sqrt{6^2+4^2}}{2}=\frac{\s qrt{52}}{2}$
$(x-1)^2+(y-5)^2=r^2$
$(x-1)^2+(y-5)^2=\frac{13}{2}$

**CaptainBlack** · December 6th 2008, 12:36 AM

Originally Posted by Arch_Stanton

$S=(\frac{x_A+x_B}{2};\frac{y_A+y_B}{2})=(\frac{4-2}{2};\frac{7+3}{2})=(1,5)$
$r=\frac{|AB|}{2}=\frac{\sqrt{6^2+4^2}}{2}=\frac{\s qrt{52}}{2}$
$(x-1)^2+(y-5)^2=r^2$
$(x-1)^2+(y-5)^2=\frac{13}{2}$

You need to provide some words describing what you are doing and why.

CB

**Chop Suey** · December 6th 2008, 01:10 AM

Originally Posted by CaptainBlack

You need to provide some words describing what you are doing and why.

CB

Originally Posted by Arch_Stanton

1. $S=(\frac{x_A+x_B}{2};\frac{y_A+y_B}{2})=(\frac{4-2}{2};\frac{7+3}{2})=(1,5)$
2. $r=\frac{|AB|}{2}=\frac{\sqrt{6^2+4^2}}{2}=\frac{\s qrt{52}}{2}$
3. $(x-1)^2+(y-5)^2=r^2$
4. $(x-1)^2+(y-5)^2=\frac{13}{2}$

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 $\frac{\sqrt{52}}{2} = \frac{\not{2}\sqrt{13}}{\not{2}}$ . Therefore, $r^2$ 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:
$(x-h)^2+(y-k)^2 = r^2$

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

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

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