Posted by **sammy** on Saturday, September 1, 2007 at 5:07pm.

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

- calc -
**Count Iblis**, Saturday, September 1, 2007 at 5:53pm
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.

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