SOLUTION: Find an equation for a circle satisfying the given conditions: The points (7, 13) and (-3, -11) are at the ends of a diameter. The answer is: (x-2)^2 + (y-1)^ = 169 I need to kn

Algebra ->  Circles -> SOLUTION: Find an equation for a circle satisfying the given conditions: The points (7, 13) and (-3, -11) are at the ends of a diameter. The answer is: (x-2)^2 + (y-1)^ = 169 I need to kn      Log On


   

Question 486044: Find an equation for a circle satisfying the given conditions:
The points (7, 13) and (-3, -11) are at the ends of a diameter.
The answer is: (x-2)^2 + (y-1)^ = 169
I need to know HOW to get this answer... I can get all the way to achieving the first aspect of the answer, but I can't seem to get the answer 169... I end up with 26?
Thank you in advance to anyone who can help.
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

First we plot the two points:



Then we draw the diameter:



The midpoint of any diameter is the center of the circle, so
we use the midpoint formula:

 



Midpoint+=+%28matrix%281%2C3%2C+++++4%2F2%2C+%22%2C%22+%2C+2%2F2%29%29+

Midpoint+=+%28matrix%281%2C3%2C+++++2+%2C+%22%2C%22%2C+1%29%29

So the center is the point (2,1)


So we plot it and label it:



Next we find the radius which is the distance from the center (2,1).
to either end of the diameter, say (7,13). We use the distance formula:

distance+=+sqrt%28%28x%5B2%5D-x%5B1%5D%29%5E2%2B%28y%5B2%5D-y%5B1%5D%29%5E2%29

distance+=+sqrt%28%287-2%29%5E2%2B%2813-1%29%5E2%29
distance+=+sqrt%285%5E2%2B12%5E2%29
distance+=+sqrt%2825%2B144%29
distance+=+sqrt%28169%29
distance+=+13

So the radius is 13.

The standard equation for a circle is

(x - h)² + (y - k)² = r²

where the center is (h,k) and the centter is r.

So wh have (h,k) = (2,1), so h=2, k=1, and r=13.

Substituting, we have:

(x - 2)² + (y - 1)² = 13²

(x - 2)² + (y - 1)² = 169



Edwin