Saturday

August 2, 2014

August 2, 2014

Posted by **alk** on Sunday, August 3, 2008 at 11:30am.

a) Determine the other x-intercept.

b) Find the equation of the circle.

Thanks for helping!

- math -
**Damon**, Sunday, August 3, 2008 at 3:48pmWell, 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)

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