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Find the equation in standard form of the circle with center at (4, −1) and that passes through the point (−4, 1).

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Answer:

The standard form of the equation of a circle with center at (h, k) and radius r is:

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

We are given that the center of the circle is (4, -1), so h = 4 and k = -1. We also know that the circle passes through the point (-4, 1), which means that the distance from the center of the circle to (-4, 1) is the radius of the circle.

The distance between two points (x1, y1) and (x2, y2) is given by the distance formula:

d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

So the radius of the circle is:

r = sqrt((-4 - 4)^2 + (1 - (-1))^2) = sqrt(100) = 10

Now we can substitute the values of h, k, and r into the standard form equation of a circle:

(x - 4)^2 + (y + 1)^2 = 10^2

Expanding the equation gives:

x^2 - 8x + 16 + y^2 + 2y + 1 = 100

Simplifying and putting the equation in standard form, we get:

x^2 + y^2 - 8x + 2y - 83 = 0

Therefore, the equation in standard form of the circle with center at (4, −1) and that passes through the point (−4, 1) is:

x^2 + y^2 - 8x + 2y - 83 = 0

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