Line m is tangent to a circle at the point 4,1 If the circle is centered at the origin, what is the y intercept of the line m?
the slope of the radius to (4,1) is ... (1 - 0) / (4 - 0) = 1/4
the slope of m (⊥ to the radius) is ... -1 / (1/4) = -4
point-slope form of m is ... y - 1 = -4 (x - 4)
slope-intercept form is ... y = -4 x + 17
x^2+y^2 = r^2
4^2 + 1^2 = r^2
r^2 = 17
so
x^2 +y^2 = 17^2
what is slope of that circle at (4,1)
2 x dx + 2 y dy = 0
dy/dx = - x/y = -4/1 = -4
that is m in
y = mx+b
y = -4x+b
put in (4,1)
1 = -4(4) + b
b = 17
so the line is
y = -4x + 17
when x = 0, y = 17
Scott's way was smarter.
thank you , Damon
To find the y-intercept of line m, we first need to determine the slope of the line.
Since line m is tangent to the circle at the point (4,1), it means that the line is perpendicular to the radius of the circle at that point. Since the circle is centered at the origin, the radius of the circle passing through (4,1) is a line connecting the origin (0,0) and (4,1).
The slope of the radius line can be calculated using the formula:
slope = (y2 - y1) / (x2 - x1)
Given that (x1, y1) = (0, 0) and (x2, y2) = (4, 1), we can substitute these values into the formula to get:
slope_r = (1 - 0) / (4 - 0) = 1/4
Since line m is perpendicular to the radius line, the slope of line m will be the negative reciprocal of the slope of the radius line. Therefore, the slope of line m is -4.
Now that we have the slope of line m, we can substitute it and the coordinates of the tangent point (4,1) into the equation of a line (y = mx + b) to solve for the y-intercept (b).
Using the coordinates (x, y) = (4, 1) and the slope (m = -4), we can substitute these values and solve for b:
1 = -4 * 4 + b
1 = -16 + b
b = 1 + 16
b = 17
Therefore, the y-intercept of line m is 17.