find the equation of circle with center on line y=3x and tangent to line x=2y at point (2,1).
I graphed the line for the circle, I graphed the tangent line. I cant tell how to find a point on the circle line thats distance from (2,1) is going to be less then the distance from that point to any other point on the tangent line.
I know (x,3x) is going to be the center. But how do I know that (x,3x) to (2,1) is more then (x,3x) to any other point on x=2y or (x,x/2).
I tried putting x,3x as the equation of a circle from 2,1 :
(2-x)^2+(1-3x)=r^2 and putting x=2y into this circle equation since x=2y == (x,x/2) and got (2-x)^2 + (1-3(x/2))^2 = r^2 but I kept going with this and got nowhere.
Im probably missing something. can someone please point me in the right direction?
Dec 19th 2010, 12:28 AM
A tangent to a circle is always perpendicular to the circle's radius.
Dec 19th 2010, 12:30 AM
Here's one way to do it. A radius is always perpendicular to a tangent so a line through the center of the circle and the point of tangency is perpendicular to the tangent line. The line x= 2y has slope 1/2 and so any perpendicular line has slope -2.
The line through (2, 1) with slope -2 has equation y= -2(x- 2)+ 1= -2x+ 4. The point where that line intersects y= 3x is the center of the circle and, of course, the distance from that point to (2, 1) is the radius.
Dec 19th 2010, 08:28 AM
thanks people, never knew that the tangent line had to be perpendicular to the center of the circle. But that made it easy. thanks!