Find an equation of the line that has a positive slope and is tangent to the circle (x-1)^2 +(y-1)^2=4 at one of its y-intercepts . Write your answer in the form of y=mx+b.
To find the equation of the line that is tangent to the circle and passes through one of its y-intercepts, we can follow these steps:
1. Find the center and radius of the circle by comparing the equation to the standard form of a circle, which is (x - h)^2 + (y - k)^2 = r^2. In this case, we have (x - 1)^2 + (y - 1)^2 = 4. So the center of the circle is (1, 1) and the radius is 2.
2. Determine the y-intercepts of the circle by substituting x = 0 into the equation. We have (0 - 1)^2 + (y - 1)^2 = 4, which simplifies to 1 + (y - 1)^2 = 4. Solving for y gives us y = 1 ± √3.
3. Since we want the line to be tangent to the circle, it should touch the circle at only one point. Therefore, we choose the y-intercept where y = 1 + √3.
4. Find the slope of the line. Since the line is tangent to the circle, it will be perpendicular to the radius drawn from the point of tangency. The radius has a slope of (y2 - y1) / (x2 - x1), where (x1, y1) is the center of the circle and (x2, y2) is the point of tangency. In this case, the center of the circle is (1, 1) and the point of tangency is (0, 1 + √3). Therefore, the slope of the radius is (1 + √3 - 1) / (0 - 1) = -√3.
5. As the line is perpendicular to the radius, the slope of the line will be the negative reciprocal of the slope of the radius. Therefore, the slope of the line is 1 / √3 = √3 / 3.
6. Now, we have the slope of the line (m = √3 / 3) and the y-intercept where y = 1 + √3. Plug these values into the slope-intercept form of the equation of a line, which is y = mx + b. Substitute the slope and the coordinates of any point (x, y) on the line:
y = (√3 / 3)x + b
7. Substituting the coordinates (0, 1 + √3) into the equation, we can solve for b:
1 + √3 = (√3 / 3)(0) + b
b = 1 + √3
8. Finally, substituting the value of b back into the equation, we get the equation of the line:
y = (√3 / 3)x + 1 + √3
Therefore, the equation of the line that has a positive slope and is tangent to the circle (x - 1)^2 + (y - 1)^2 = 4 at one of its y-intercepts is y = (√3 / 3)x + 1 + √3.