Equation of Circle Tangent to Y-Axis, X-Axis and Line

**juankim111222** · June 8th 2010, 02:07 PM

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.

**tonio** · June 8th 2010, 03:43 PM

Originally Posted by juankim111222

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.

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

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

Tonio

**juankim111222** · June 8th 2010, 04:52 PM

Thanks. So is there any way to find the center $(r,r)$ if the only information given is the tangent line $3x+4y-12=0$ ?

**Archie Meade** · June 8th 2010, 05:35 PM

Originally Posted by juankim111222

Thanks. So is there any way to find the center $(r,r)$ if the only information given is the tangent line $3x+4y-12=0$ ?

Just find where the line crosses the x and y axes.

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

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

Then the triangle area is 2(3)=6

and the hypotenuse from Pythagoras' theorem is 5.

Then the radius is

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

Hence, the centre is (1,1) and the radius is 1.

**tonio** · June 8th 2010, 06:05 PM

Originally Posted by juankim111222

Thanks. So is there any way to find the center $(r,r)$ if the only information given is the tangent line $3x+4y-12=0$ ?

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 $(r,r)$ to

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

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

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

Tonio

Ps. There's a mistake above: $r=6$ is impossible (why?)

**Archie Meade** · June 8th 2010, 06:59 PM

Originally Posted by juankim111222

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.

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...

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

$a+b-2r=c$

$2r=a+b-c$

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

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

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.

**yvonne1981** · May 12th 2012, 01:11 PM

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?

**earboth** · May 12th 2012, 11:37 PM

Originally Posted by tonio

...

Tonio

Ps. There's a mistake above: $r=6$ is impossible (why?)<-- no

Originally Posted by Archie Meade

...

Tonio's method finds both solutions.
the circle of r=6 is also tangent to all 3 lines.

Originally Posted by yvonne1981

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? <--- see Tonio's post

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.

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