The circle is tangent to the y-axis at y = 3. From this it follows that the center of the circle must be located somewhere on the line y = 3. Therefore b = 3.
The point (0,3) is on the circle, so:
a^2 = r^2
The point (1,0) is on the circle:
(1-a)^2 + 9 = r^2
Suntracting these two equations gives:
a^2 - (1-a)^2 = 0 ---->
(2a - 1) =0---->
a = 1/2
And for this you can easily find r.
a circle is tangent to the y-axis at y=3 and has one x-intercept at x=1
1. determine the other x-intercept
2. find an equation for the circle
so i know two points on the circle are (0,3) and (1,0) and the equation for a circle is (x-a)^2 + (y-b)^2 = r^2 but i don't know how to solve the two questions
1 answer
1 answer