Well, if it is tangent to the y axis at (0,3), then the center must be somewhere on the horizontal line y = 3 because the radius is perpendicular to the tangent.
So if center is at (a,3)
(x-a)^2 + (y-3)^2 = r^2
we know when x = 0, y = 3 so
a^2 = r^2
so
(x-a)^2 + (y-3)^2 = a^2
now when y = 0, x = 1 and something else
(1-a)^2 +9 = a^2
1 -2a +a^2 +9 = a^2
a = 5 = r
so center is at (5,3)
(x-5)^2 + (y-3)^2 = 25
is the equation
x intercepts:
(x-5)^2 +9 = 25
x^2 - 10 x + 25 + 9 = 25
x^2 -10 x = -9
x^2 - 10 x + 25 = 16
(x-5)^2 = 16
x-5 = +/- 4
x = 1 or x = 9
which you could have seen from a sketch once you knew the radius was 5 and the center at (5,3)
A circle is tangent to the y-axis at y=3 and has one intercept at x=1.
a) Determine the other x-intercept.
b) Find the equation of the circle.
Thanks for helping!
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