# Math Help - help please

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

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

3. 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

4. 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:
$(x-1)^2+(y-5)^2 = 13$