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Find an equation of the circle whose diameter has endpoints (2,-4) and (-6,4)

User NFDream
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Answer:

(x + 1)^2 + y^2 = 64

Explanation:

Step 1: Find the center of the circle (h, k) by taking the average of the x-coordinates and the y-coordinates of the endpoints of the diameter:

Center (h, k) = ((2 + (-6)) / 2, (-4 + 4) / 2)

Center (h, k) = (-2 / 2, 0 / 2)

Center (h, k) = (-1, 0)

So, the center of the circle is (-1, 0).

Step 2: Calculate the radius of the circle (r), which is half the length of the diameter. Use the distance formula between the two endpoints of the diameter:

Distance formula: d = √((x2 - x1)^2 + (y2 - y1)^2)

Diameter length = √((-6 - 2)^2 + (4 - (-4))^2)

Diameter length = √((-8)^2 + (8)^2)

Diameter length = √(64 + 64)

Diameter length = √128

Radius (r) = Diameter length / 2

Radius (r) = √128 / 2

Radius (r) = √64

Radius (r) = 8

Step 3: Now that we have the center (h, k) and the radius (r), we can use the standard form of the equation of a circle:

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

Substitute the values of the center and radius into the equation:

(x - (-1))^2 + (y - 0)^2 = 8^2

(x + 1)^2 + y^2 = 64

So, the equation of the circle whose diameter has endpoints at (2, -4) and (-6, 4) is:

(x + 1)^2 + y^2 = 64

User Jahnold
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