# find the general form of the equation of each circle

• February 18th 2010, 05:36 PM
alessandromangione
find the general form of the equation of each circle
with the endpoints of a diameter at (1,4) (-3,2)

so for the radius i have to find the distance between these two points...and i got square root of 20...then..?

hope u guys can help me

thanks
• February 18th 2010, 05:47 PM
skeeter
Quote:

Originally Posted by alessandromangione
with the endpoints of a diameter at (1,4) (-3,2)

so for the radius i have to find the distance between these two points...and i got square root of 20...then..?

hope u guys can help me

thanks

$\sqrt{20}$ is the diameter, not the radius.

the midpoint between the two points is the circle's center, $(h,k)$

$(x-h)^2 + (y-k)^2 = r^2$
• February 18th 2010, 06:08 PM
alessandromangione
Quote:

Originally Posted by skeeter
$\sqrt{20}$ is the diameter, not the radius.

the midpoint between the two points is the circle's center, $(h,k)$

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

so the the radiuus is square root of 10?
• February 18th 2010, 06:14 PM
skeeter
Quote:

Originally Posted by alessandromangione
so the the radiuus is square root of 10?

I have to believe that you were taught better than that ...

$\frac{\sqrt{20}}{2} \ne \sqrt{10}$

$\frac{\sqrt{20}}{2} = \frac{\sqrt{20}}{\sqrt{4}} = \sqrt{\frac{20}{4}} = \sqrt{5}
$