Thread: more circle problems

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?

A tangent to a circle is always perpendicular to the circle's radius.

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.

thanks people, never knew that the tangent line had to be perpendicular to the center of the circle. But that made it easy. thanks!

Hello, frankinaround!

Another approach . . .

$\displaystyle \text{Find the equation of circle with center on line }y\,=\,3x$
$\displaystyle \text{and tangent to line }x\,=\,2y\text{ at point }(2,1).$

The distance from $\displaystyle (2,1)$ to the line $\displaystyle 3x - y \:=\:0$

. . is given by: .$\displaystyle d \:=\:\dfrac{|3(2) - 1(1)|}{\sqrt{3^2+1^2}} \;=\;\dfrac{5}{\sqrt{10}} \:=\:\sqrt{\dfrac{5}{2}}\quad \hdots\;\text{This is the radius..}$


The point $\displaystyle (x,\,3x)$ which is $\displaystyle \sqrt{\frac{5}{2}}$ units from $\displaystyle (2,1)$

. . is given by: .$\displaystyle \sqrt{(x-2)^2 + (3x-1)^2} \:=\:\sqrt{\frac{5}{2}}$


Square: .$\displaystyle x^2 - 4x + 4 + 9x^2 - 6x + 1 \:=\:\frac{5}{2}$

. . which simplifies to: .$\displaystyle 4x^2 - 4x + 1 \:=\:0$

. . which factors: .$\displaystyle (2x-1)^2 \:=\:0$

. . and has one root: .$\displaystyle x \,=\,\frac{1}{2}$

Hence: .$\displaystyle y \:=\:3(\frac{1}{2}) \:=\:\frac{3}{2}$


The center is $\displaystyle \left(\frac{1}{2},\,\frac{3}{2}\right)$ and the radius is $\displaystyle \sqrt{\frac{5}{2}}$

The equation of the circle is: .$\displaystyle \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 \;=\;\frac{5}{2}$