Question

Write the equation of a circle whose center is at (7,5) and contains the points (3,-2)

Answer 1

Well i didn't know if i could solve it but i thought, if you wanted to make a circle and that it touches that point, then the point would be the radius, so to find the radius it would be the distance between those two points. Using the formula *sqrt((x2 - x1)^2 + (y2 - y1)^2
now we rewrite it but im going to use the points backwards since it works easier that way;
*sqrt((7 - 3)^2 + (5 -(-2)^2
*sqrt((4)^2 + (5 +2)^2
*sqrt((4)^2 + (7)^2
*sqrt(16 + 49
*sqrt(64
So the radius is equal to the *sqrt(64
Now using the equation of a circle, (x -h)^2 + (y - k)^2 = r^2
where (h,k) is the center and r is the radius.

(x - (7))^2 + (y - (5))^2 = (*sqrt(64)^2
(x - 7)^2 + (y - 5)^2 = 64 ---------- since *sqrt of 64 times itself is equal to 64

so thats the answer, "(x - 7)^2 + (y - 5)^2 = 64 "; if anything you can check it out by using a graphing technology

Answer 2

wait, i meant 65 --------- where their was the 64's it was a 65. Sorry i wrote it wrong