What is the minimum distance between any point on the circle x^2 + y^2 = 25 and the line y = -\frac{3}{4}x + \frac{75}{4} ?
The distance from (h,k) to the line ax+by+c=0 is |ah+bk+c|/√(a^2+b^2)
y = 3/4 x + 75/4 is
3x - 4y + 75 = 0
we know that if (h,k) is on the circle,
k=√(25-h^2)
so, the distance s from (h,k) to 3x-4y+75=0 is
s = (3h-4√(25-h^2)+75)/5
find where s'=0 and that will give you h, and you can figure k.
Or, you know that at (h,k) on the circle, the slope is -h/k.
The slope of the line is 3/4
The minmum distance will be a line perpendicular, over to the circle.
So, we want a line with slope -4/3 = -h/k, and we want it in QII, since that's where the line is.
Looks like (-4,3) is the point we want.
Just for grins, run through the other solution and see whether it agrees.
To find the minimum distance between a point on the circle and a line, we need to find the perpendicular distance between them.
First, let's rewrite the equation of the line in the standard form: y = mx + b, where m is the slope and b is the y-intercept.
For the given line y = -\frac{3}{4}x + \frac{75}{4}, the slope (m) is -\frac{3}{4} and the y-intercept (b) is \frac{75}{4}.
Now, let's find the equation of the line perpendicular to the given line. The slope of a line perpendicular to a line with slope m is the negative reciprocal of m. So, the slope of the perpendicular line is \frac{4}{3}.
Next, let's find the intersection point between the given line and the perpendicular line. To do this, we need to solve the system of equations formed by the line and the circle.
Substituting the equation of the line into the equation of the circle, we get:
x^2 + (-\frac{3}{4}x + \frac{75}{4})^2 = 25
Simplifying this equation, we can solve for x:
x^2 + (-\frac{3}{4}x)^2 + 2(-\frac{3}{4}x)(\frac{75}{4}) + (\frac{75}{4})^2 = 25
x^2 + \frac{9}{16}x^2 - \frac{135}{8}x + \frac{5625}{16} = 25
\frac{25}{16}x^2 - \frac{135}{8}x + \frac{5625}{16} - 25 = 0
\frac{25}{16}x^2 - \frac{135}{8}x + \frac{575}{16} = 0
This quadratic equation can be solved using the quadratic formula: x = (-b ± sqrt(b^2 - 4ac))/(2a).
Using this formula, we find the solutions for x:
x = (-(-\frac{135}{8}) ± sqrt((-frac{135}{8})^2 - 4 * (\frac{25}{16}) * (\frac{575}{16})) / (2 * (\frac{25}{16}))
After simplifying, we get two possible values for x. Let's call them x1 and x2.
Now, substitute these values of x into the equation of the line to find the corresponding y-coordinates. Let's call them y1 and y2.
Next, we calculate the perpendicular distance from the center of the circle (0, 0) to each of these two points (x1, y1) and (x2, y2) using the distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
The smaller of these two distances will be the minimum distance between any point on the circle and the given line.