Acircle of center (a, o) passes through the point (o, -a) find it equation

  (x -a)^2 + y^2 = 2a^2

Step-by-step explanation:

The equation of a circle with center (h, k) and radius r is ...

  (x -h)^2 +(y -k)^2 = r^2

Here r^2 is the square of the distance between the center and the point on the circle. That can be found using the distance formula:

  d^2 = r^2 = (x2-x1)^2 +(y2 -y1)^2 = (0-a)^2 +(-a-0)^2 = 2a^2

Filling in the above formula with (h, k) = (a, 0) and r^2 = 2a^2, we find the equation to be ...

  (x -a)^2 +y^2 = 2a^2

answer

the correct answer is

explanation

the given function is

the excluded values are obtained by solving:

we factor to obtain:

the excluded values are

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