find an equation of the circle satisfying the given condition

Question

center (5,0)

passes through the point (5,-4)

Answer 1

Center at (5,0).  Radius 4.

Equation of a circle:

(x-a)^2 + (y-b)^2 = r^2

where center is at (a,b) and radius = r

(x-5)^2 + y^2 = 16

Answer 2

????????????? "passessseseseesss thru" ?????

it mite GO THRU a point.

(5, -4) be 4 yunits from senter=(5,0)

won wae wood be tu sae radius=4, so kerv be horizontal wen go thru (5, -4)

but htat oenlee be 1 av infinut maebees

Comments

The distance from the center to any point on a circle is the radius, so if the circle is centered on (5,0) and passes through (5,-4), the radius must be 4.

With a center of (5,0) and a radius of 4, the circle is uniquely defined.