Please help algebra 2

Question

Write the equation in standard form for the circle passing through (-8,2) centered at the origin.

Answer 1

The equation of a circle centered at point (h,k) having radius r is

(x-h)² + (y-k)² = r²

In our case, the circle is centered at the origin (0,0). Therefore, h = k = 0. So, the equation becomes

x² + y² = r²

Now, in order to find the radius, we use the fact that the circle passes through point (-8,2). The above equation must be satisfied for x = -8, y = 2. Therefore

(-8)² + (2)² = r²

or

r² = 68

Therefore, the equation of the given circle is

x² + y² = 68

Answer 2

The form of an equation of a circle with center (h,k) and radius r, is (x-h)² + (y-k)² = r². So the equation you want has a radius that connects (0,0), the origin, and (-8,2). Use the distance formula to find this distance, which is the square root of 68 or about 8.25. So the formula for your circle would be (x-0)² + (y-0)² = 8.25² or x² + y² = 68. Hope it helps Parker!!