SOLUTION: A circle of radius 6 touches both coordinate axes. A line with slope -3/4 passes over and just touches the circle. If the circle is in the first quadrant, find the equation of the

Question 1148246: A circle of radius 6 touches both coordinate axes. A line with slope -3/4 passes over and just touches the circle. If the circle is in the first quadrant, find the equation of the line.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Since the circle is in the first quadrant and touches (is tangent to) both coordinate axes, the center of the circle is (6,6).

Let P be the point of tangency of the line and circle.

The line has slope -3/4, so the radius to the point of tangency has a slope of 4/3. Then, given that the radius of the circle is 6, the Pythagorean Theorem can be used to determine that P is 3.6 units to the right of and 4.8 units above the center of the circle. So the coordinates of P are (6+3.6,6+4.8) = (9.6,10.8).

You now have the slope of the line and a point on the line; there are many ways to get from there to an equation of the line. I leave that much of the problem for you....

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