Need help please:

(3,0) (7,6)
(x-7)2 + (y-6)2 = 7(2)
(x-7)2 + (y-6)2 = 49

Having difficulty with this problem.Can someone help?

geometry - Reiny, Saturday, March 31, 2012 at 8:07am
You don't say what the actual problem is.

I see the resemblance to finding the equation of a circle??

What are you trying to find?

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geometry - Kel, Saturday, March 31, 2012 at 6:59pm
It says to write an equation of a circle with diameter
__
AB

Is A at (3,0) and B at (7,6) ?

if so then the center is halfway between.
Xc = (3+7)/2 = 5
Yc = (0+6)/3 = 3
so of form
(x - 5)^2 + (y - 3)^2 = r^2
but r is half the length of AB
AB^2 = 4^2 + 6^2 = 16+36 = 52
AB = 7.21
(1/2) AB = r = 3.61
r^2 = 13
so in the end
(x - 5)^2 + (y - 3)^2 = 13

6 = x + 2

To write the equation of a circle given the diameter, you need to know the coordinates of the endpoints of the diameter. In this case, the endpoints of the diameter are (3,0) and (7,6).

To find the center of the circle, you can find the midpoint of the diameter. The midpoint formula is given by:

Midpoint = ((x1 + x2)/2, (y1 + y2)/2)

Plugging in the coordinates of the endpoints, we get:

Midpoint = ((3 + 7)/2, (0 + 6)/2)
= (5, 3)

So the center of the circle is (5,3).

The radius of the circle can be found by taking half the length of the diameter. The distance formula is given by:

Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Plugging in the coordinates of the endpoints, we get:

Distance = sqrt((7 - 3)^2 + (6 - 0)^2)
= sqrt(16 + 36)
= sqrt(52)

So the radius of the circle is sqrt(52), which can be simplified to 2sqrt(13).

The equation of a circle with center (h,k) and radius r is given by:

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

Plugging in the values, we get:

(x-5)^2 + (y-3)^2 = (2sqrt(13))^2
(x-5)^2 + (y-3)^2 = 52

So the equation of the circle with diameter AB is (x-5)^2 + (y-3)^2 = 52.