Question

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

Answer 1

The equation in standard form for the circle passing through (2,6) centered at the origin is x² + y² = 40.

Step-by-step explanation:

To write the equation of a circle in standard form, we use the formula (x - h)² + (y - k)² = r², where (h, k) represents the center of the circle and r represents the radius.

In this case, the center of the circle is the origin (0, 0) and the circle passes through the point (2, 6). So, the radius of the circle is the distance from the origin to (2, 6), which is √(2² + 6²) = √40 = 2√10.

Substituting the values into the formula, we get the equation as (x - 0)² + (y - 0)² = (2√10)².

Answer 2

Answer: x^2 + y^2 = 40

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Step-by-step 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.

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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