# more circle problems

• December 19th 2010, 12:25 AM
frankinaround
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?
• December 19th 2010, 12:28 AM
Prove It
A tangent to a circle is always perpendicular to the circle's radius.
• December 19th 2010, 12:30 AM
HallsofIvy
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.
• December 19th 2010, 08:28 AM
frankinaround
thanks people, never knew that the tangent line had to be perpendicular to the center of the circle. But that made it easy. thanks!
• December 19th 2010, 11:35 AM
Soroban
Hello, frankinaround!

Another approach . . .

Quote:

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

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

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

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

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

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

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

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

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

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

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

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