how to write an equation of a circle given the center points and the points it passes through?

Answer 1

kind of like the first guy but better.
im guessing that you have the center and then a point that it passes through so you need to use the distance formula to find the radius
r = sqrt((x-x1)^2 + (y-y2)^2)
then follow the first person

Answer 2

The equation of a circle is (x - h)^2 + (y - ok)^2 = r^2. If the middle is on the starting place, then h and ok are 0. accordingly, a. If there's a aspect (5, 0) on the circle, the radius is 5 and the equation will change into x^2 + y^2 = 25. b. If there's a aspect (2, 0) on the circle, the radius is two and the equation will change into x^2 + y^2 = 4.

Answer 3

The general equation of a circle

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

(h,k) = center
(x,y) = a point that the circle passes through.

For ex:

Center : (4,5)
Passes through : (1,2)

(1 - 4)^2 + (2 - 5)^2 = r^2
(-3)^2 + (-3)^2 = r^2
9 + 9 = r^2
18 = r^2

so your equation would be

(x - 4)^2 + (y - 5)^2 = 18

Answer 4

Given the center point (x_center, y_center) and the radius r.

(x-x_center)^2 + (y-y_center)^2 = r^2

If centered at (1,3) with radius of 2, then you have:
(x-1)^2 + (y-3)^2 = 4

Answer 5

the equation of the circle with centre h and k and radius r is
(x-h)^2+(y-k)^2=r^2
now substitute the x and y coordinates of the point for x and y and h and k &r will trickle out

Answer 6

say the center is (a,b)
and it passes through (z,w)
r=sqrt ( (z-a)^2 + (w-b)^2 )

then
(x-a)^2 +(y-b)^2 =r^2