Write an equation for a circle with a diameter that has endpointsat (-4, -7) and (-2 ,-5)

(x + 3)^2 + (y + 6)^2 = (√2)^2 = 2

Step-by-step explanation:

The standard equation for a circle with center at (h, k) and radius r is

(x - h)^2 + (y - k)^2 = r^2.  We need to find h, k and r for this particular circle.

The length of the radius is half the length of the diameter.  Find the length of this diameter first, and than halve the result.

diameter^2 = 2^2 + 2^2, or diameter^2 = 8, or diameter = √8 = 2√2.

This is because x changes from -4 to -2 (+2) and y changes from -7 to -5 (+2).

Then the radius is r = 2√2.

The coordinates of the center are found using the midpoint formula:

                 -4-2

xcenter = = -3

                    2

                 -7-5

ycenter = = - 6

                     2

Inserting -3 for h, -6 for k and √2 for r, we get:

(x + 3)^2 + (y + 6)^2 = (√2)^2 = 2

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.

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