Find center on y=1/3x+6, tangent to y=x+20(11;9)
hmmm. I see a couple of straight lines, and what may be a point. I fail to see any circles or other curves involved.
Maybe a little more explanation will help.
Do you want the equation of a circle whose center is on y=1/3x+6, and is tangent to y=x+20?
If so, where does (11,9) come in?
the given is 0.33x+16, tangent to y=x+20(11,9)
yeah, you already said that. But what does
tangent to y=x+20(11,9)
mean?
To find the center of a circle that is tangent to a line, you need to first identify the point of tangency. In this case, the line y = x + 20 is tangent to the circle at the point (11, 9).
To proceed, we will use the fact that the radius of the circle is perpendicular to the tangent line at the point of tangency. The slope of the tangent line is 1, so the slope of the radius will be -1 (perpendicular).
Since the center of the circle lies on the line y = (1/3)x + 6, we can write the equation of the line perpendicular to it as y = -3x + b (using point-slope form).
Now we can plug in the coordinates of the point (11, 9) to find the corresponding value of b.
9 = -(3)11 + b
9 = -33 + b
b = 42
Therefore, the equation of the line perpendicular to y = (1/3)x + 6 and passing through the point (11, 9) is y = -3x + 42.
Now we can solve the system of equations formed by the two lines to find their point of intersection, which will be the center of the circle.
Substituting y = -3x + 42 into y = (1/3)x + 6, we get:
-3x + 42 = (1/3)x + 6
-3x - (1/3)x = 6 - 42
-10/3x = -36
x = (-36) / (-10/3)
x = 36/10 * 3/1
x = 108/10
x = 54/5
Now substituting this value back into either equation, y = (1/3)x + 6, we get:
y = (1/3)(54/5) + 6
y = 18/5 + 6
y = (18+30) / 5
y = 48/5
Therefore, the center of the circle is at the point (54/5, 48/5).