Question 1143730
Here is a VERY DIFFERENT solution method, no better and no worse than the one shown by tutor @ikleyn.  It shows nicely how two completely different methods can be used to solve a problem, if you have the tools you need.<br>
It should be noted that this method is possible for this example only because the triangle is a right triangle with one side on the x-axis.  There are no such restrictions on the general method shown by @ikleyn.<br>
The equations of the lines containing the sides of the triangle are
(1) {{{y = 0}}}
(2) {{{y = (-4/3)x+50/3}}}
(3) {{{y = (3/4)x}}}<br>
The slopes of (2) and (3) show that the triangle is a right triangle; and the slope of (3) tells us that the right triangle has side lengths in the ratio 3:4:5.<br>
The x-intercept of line (2) is (12.5,0).  That means the hypotenuse of the triangle is 12.5.  And that, along with the known 3:4:5 ratio of the side lengths, tells us the lengths of the two legs are 7.5 and 10.<br>
So the perimeter of the triangle is 7.5+10+12.5 = 30; and the area is (1/2)(7.5)(10) = 75/2.<br>
But the area is also one-half the perimeter times the radius of the inscribed circle, so we can find the radius of the circle:
{{{(1/2)(30)r = 75/2}}}
{{{15r = 75/2}}}
{{{r = 5/2}}}<br>
So we have the radius of the circle; now we need to find the center, (h,k).<br>
Since the radius is 5/2 and one side of the triangle is on the line y=0, we know the y coordinate of the center is 5/2.  So the center of the circle is (h,5/2).<br>
To find the x coordinate of the center, we can use the fact that the measure of the angle between the x-axis and the line from the origin to the center of the circle is half the measure of the angle of the triangle at the origin.<br>
{{{tan(theta/2) = sqrt((1-cos(theta))/(1+cos(theta)))}}}
{{{tan(theta/2) = k/h = sqrt((1-4/5)/(1+4/5)) = sqrt((1/5)/(9/5)) = sqrt(1/9) = 1/3}}}<br>
So<br>
{{{k/h = (5/2)/h = 1/3}}}
{{{h = 15/2}}}<br>
We now have the radius of the circle and the coordinates of the center of the circle, so we can write the equation of the circle:<br>
center(15/2,5/2), radius 5<br>
{{{(x-15/2)^2+(y-5/2)^2 = (5/2)^2}}}