SOLUTION: Find the equation of the circle with center at (6,3) and passes through the point (-6,-1)

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Question 1153880: Find the equation of the circle with center at (6,3) and passes through the point (-6,-1)
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-6%29%5E2%2B%28y-3%29%5E2=%28-6-6%29%5E2%2B%28-1-3%29%5E2
Do what you can from that.



---MORE DETAIL----
Basic equation for circle in coordinate plane: %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2
r is the radius.
point (h,k) is center point of circle.

Your given center is (6,3).
Point on circle is given as (-6,-1).

Left member of equation you want is %28x-6%29%5E2%2B%28y-3%29%5E2.
Use the given point ON the circle to find the value of r%5E2, the right member of the equation.

%28x-6%29%5E2%2B%28y-3%29%5E2=%28-6-6%29%5E2%2B%28-1-3%29%5E2
AND DO THE NECESSARY COMPUTATION to get this value for r%5E2;
%28-12%29%5E2%2B%28-4%29%5E2
144%2B16
160

Finished equation is highlight%28%28x-6%29%5E2%2B%28y-3%29%5E2=160%29.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2 <---> center (h,k), radius r

Example: center (3,-2) and radius 5 --> %28x-3%29%5E2%2B%28y%2B2%29%5E2+=+25

Example: %28x%2B1%29%5E2%2B%28y-4%29%5E2+=+16 --> center (-1,4), radius 4

You are given the center (h,k).

The radius r is the distance from the center to the other given point. Use the distance formula (AKA Pythagorean Theorem) to determine r.

Then plug your h, k, and r in the formula.