Question

Which equation represents a circle that contains the point (–2, 8) and has a center at (4, 0)? Distance formula: StartRoot (x 2 minus x 1) squared + (y 2 minus y 1) squared EndRoot (x – 4)2 + y2 = 100 (x – 4)2 + y2 = 10 x2 + (y – 4)² = 10 x2 + (y – 4)² = 100

Answer 1

Answer:

The answer to your question is:    (x - 4)² + (y)² = 100

Step-by-step explanation:

Data

Circle

Center = (4, 0)

Point = ( -2, 8)

Equations

distance = √(x2 - x1)² + (y2 - y1)²

Circle   (x - h)² + (y - k)² = r²

Process

Find r

                r  = √(x2 - x1)² + (y2 - y1)²

                r =  √(-2 - 4)² + (8 - 0)²

                r = √ (-6)² + (8)²

                r = √ 36 + 64

                r = √ 100

                r = 10

Circle

               (x - 4)² + (y - 0)² = (10)²

               (x - 4)² + (y)² = 100

Answer 2

The equation of the given circle will be given as:

$(x-4)^2 + y^2 = 100$

Thus, Option B is correct.

Given that:

• A point on the circle :  P(-2,8)
• Coordinates of center of the circle: (4,0)

The equation of a circle with center at (h,k) is given as:

$(x-h)^2 + (y-k)^2 = r^2$, where r = radius of the circle.

Since P(-2,8) is a point on the circle, thus it must satisfy the equation of the given circle.

$(x-h)^2 + (y-k)^2 = r^2\\(-2-4)^2 + (8-0)^2 = r^2\\36 + 64 = r^2\\100 = r^2\\10 = r \: \: \text{since r cannot be negative as radius is measured in non-negative units}$

Thus, equation of the given circle will be given as:

$(x-4)^2 + y^2 = 100$

The graph of that circle along with given point on it and center of the circle is attached below.

Learn more about equation of circle here:

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