C (3,4) tangent to the line y=1/3x-1/3
I do not understand what you are asking.
I wanna know the radius of the circle
The distance from a point (h,k) to the line Ax^2+Bx+C=0 is
|Ah+Bk+C|/√(A^2+B^2)
We can avoid messy fractions if we rewrite the line as
x-3y-1=0
So, that means the radius of your circle is
|1*3-3*4-1|/√(1^2+3^2) = 10/√10 = √10
Now you know the center and the radius, so just put into standard circle form.
Thank you
To determine if the point (3, 4) is tangent to the line y = (1/3)x - (1/3), we can follow these steps:
1. Substitute the x and y coordinates of the given point (3, 4) into the equation of the line and check if they satisfy the equation.
y = (1/3)x - (1/3)
4 = (1/3)(3) - (1/3)
4 = 1 - (1/3)
4 = (3/3) - (1/3)
4 = 2/3
Since 4 does not equal 2/3, the point (3, 4) is not on the line.
2. Calculate the slope of the line y = (1/3)x - (1/3).
The slope-intercept form of a line is y = mx + b, where m represents the slope.
Comparing the given line's equation, y = (1/3)x - (1/3), to the slope-intercept form, we can see that the slope is 1/3.
3. Calculate the slope between the point (3, 4) and the line.
The slope between two points (x1, y1) and (x2, y2) is given by the formula m = (y2 - y1) / (x2 - x1).
Let's consider the point (x1, y1) = (3, 4) and a point (x2, y2) on the line. We can take any convenient value for x2 and calculate the corresponding y2.
Let's choose x2 = 0, which gives us:
y2 = (1/3)(0) - (1/3)
y2 = 0 - (1/3)
y2 = -1/3
Now we can calculate the slope: m = (-1/3 - 4) / (0 - 3)
m = (-4 1/3) / (-3)
m = (13/3) / (3/1)
m = 13/9
4. Compare the slopes calculated in steps 2 and 3.
The slope of the line is 1/3, and the slope between the point (3, 4) and the line is 13/9.
Since the slopes are not equal, the point (3, 4) is not tangent to the line y = (1/3)x - (1/3).
Therefore, the point (3, 4) is not tangent to the line y = (1/3)x - (1/3).