Please help me with the process of finding the equation of a circle in the first quadrant that is tangent to the x-axis, y-axis, and a line of the form ax+by+c=0. Thanks.

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- Jun 8th 2010, 02:07 PMjuankim111222Equation of Circle Tangent to Y-Axis, X-Axis and Line
Please help me with the process of finding the equation of a circle in the first quadrant that is tangent to the x-axis, y-axis, and a line of the form ax+by+c=0. Thanks.

- Jun 8th 2010, 03:43 PMtonio

A circle $\displaystyle (x-a)^2+(y-b)^2=r^2$ is tangent to the x-axis iff $\displaystyle |b|=r$ , is tangent to the y-axis iff $\displaystyle |a|=r$ , and since the circle's center is in the first quadrant this means that the circle's equation is $\displaystyle (x-r)^2+(y-r)^2=r^2$ , and center at $\displaystyle (r,r)$

Finally, the distance from a given point $\displaystyle (x_1,y_1)$ to a given line $\displaystyle Ax+By+C=0$ is given by $\displaystyle \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$ . Do some maths now.

Tonio - Jun 8th 2010, 04:52 PMjuankim111222
Thanks. So is there any way to find the center $\displaystyle (r,r)$ if the only information given is the tangent line $\displaystyle 3x+4y-12=0$?

- Jun 8th 2010, 05:35 PMArchie Meade
Just find where the line crosses the x and y axes.

$\displaystyle y=0\ \Rightarrow\ 3x-12=0\ \Rightarrow\ x=4$

$\displaystyle x=0\ \Rightarrow\ 4y-12=0\ \Rightarrow\ y=3$

Then the triangle area is 2(3)=6

and the hypotenuse from Pythagoras' theorem is 5.

Then the radius is

$\displaystyle r=\frac{2A}{sum\ of\ the\ side\ lengths}=\frac{2(6)}{5+4+3}=1$

Hence, the centre is (1,1) and the radius is 1. - Jun 8th 2010, 06:05 PMtonio

Do you mean a tangent circle to the two axis and also to that equation? If so input the values in the equation for the distance of the point $\displaystyle (r,r)$ to

the line $\displaystyle 3x+4y-12=0$:

$\displaystyle \frac{|3r+4r-12|}{\sqrt{3^2+4^2}}=r\iff 49r^2-168r+144=$ $\displaystyle 25r^2\iff r^2-7r+6=0\Longrightarrow r=1,\,6$ , so there are two possibilities:

$\displaystyle (x-1)^2+(y-1)^2=1\,,\,\,\,(x-6)^2+(y-6)^2=36$

Tonio

Ps. There's a mistake above: $\displaystyle r=6$ is impossible (why?) - Jun 8th 2010, 06:59 PMArchie Meade
if the circle is only in the first quadrant, here is another way to do it.

See the attachment...

the triangle contains pairs of identical right-angled triangles,

hence...

$\displaystyle a-r+b-r=c$ where c is the length of the hypotenuse, the 3rd side of the triangle.

$\displaystyle a+b-2r=c$

$\displaystyle 2r=a+b-c$

$\displaystyle r=\frac{a+b-c}{2}$

$\displaystyle c=\sqrt{a^2+b^2}=\sqrt{3^2+4^2}=5$

Then it is straightforward to find r.

Tonio's method finds both solutions.

the circle of r=6 is also tangent to all 3 lines. - May 12th 2012, 01:11 PMyvonne1981Re: Equation of Circle Tangent to Y-Axis, X-Axis and Line
What about the circle to the right of the line, that is also in the first quadrant and tangent to the line and both axes? how do we find its radius?

- May 12th 2012, 11:37 PMearbothRe: Equation of Circle Tangent to Y-Axis, X-Axis and Line
In addition to the previous posts I've done a geometrical construction: Yo'll find the two centers as Points of intersection of two of the three angle bisectors.

The colour of the bisector correspond to the colour of the circle. All three angle bisectors form a right triangle.