Each point on the edge of a circle is equidistant from the center of the circle.
The center of a circle is located at (6, 3). Which point on the y-axis could be
on the edge of the circle if the distance from the center of the circle to the
edge is 10 units?
This describes a circle with center (h,k) = (6,3) and radius r = 10
That circle has the equation
(x-h)²+(y-k)² = r²
(x-6)²+(y-3)² = 10²
(x-6)²+(y-3)² = 100
Every point on the y-axis has 0 as its x-coordinate, so we
substitute x=0 in
(x-6)²+(y-3)² = 100
(0-6)²+(y-3)² = 100
(-6)²+(y-3)² = 100
36+(y-3)² = 100
(y-3)² = 64
y-3 = ±8
y = 3±8
y=3+8, y=3-8
y=11, y=-5
Two solutions: (0,11) and (0,-5)
Edwin