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
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.
To determine the other x-intercept of the circle, you can use the fact that the circle is tangent to the y-axis at y = 3. Since a tangent line is perpendicular to the radius of the circle at the point of tangency, the center of the circle must lie on the line y = 3. Thus, the y-coordinate of the center is 3.
We already know that one x-intercept is at x = 1, which means the distance between the center of the circle and this x-intercept is equal to the radius. From the given points (0, 3) and (1, 0), we can find the distance between them using the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((1 - 0)^2 + (0 - 3)^2)
= sqrt(1^2 + (-3)^2)
= sqrt(1 + 9)
= sqrt(10)
Since the radius is equal to this distance, the radius of the circle is √10.
Now, we can write the equation for the circle using the center and radius. The center of the circle is (a, b) = (0, 3), and the radius is r = √10. Plugging these values into the standard equation for a circle, we have:
(x - a)^2 + (y - b)^2 = r^2
(x - 0)^2 + (y - 3)^2 = (√10)^2
x^2 + (y - 3)^2 = 10
This is the equation for the circle. To find the other x-intercept, you can substitute y = 0 into the equation and solve for x:
x^2 + (0 - 3)^2 = 10
x^2 + 9 = 10
x^2 = 1
x = ±1
So, the other x-intercept of the circle is x = -1.