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 $\displaystyle (x-a)^2+(y-b)^2=r^2$ 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
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.
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?)
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.
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.