Write the equation in standard form for the circle passing through (2,6) centered at the origin.

Explanation:

The center is the origin, so (h,k) = (0,0).

The point (x,y) = (2,6) is on the circle's edge. We'll use these four values to find the value of r^2

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

(2-0)^2 + (6-0)^2 = r^2

2^2 + 6^2 = r^2

4 + 36 = r^2

40 = r^2

r^2 = 40

We don't need to solve for r itself. If you wanted to, you'd get the radius to be r = sqrt(40) = 2*sqrt(10) = 6.324555 approximately.

Since h = 0, k = 0, and r^2 = 40, we can say:

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

(x-0)^2 + (y-0)^2 = 40

x^2 + y^2 = 40

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.