well, you know if it is tangent to both axis, the center is at r,-r
It passes through 8,-9 so the distance from that point to the center is also r.
r^2=(8-r)^2 + (-9+r)^2
= 64-16r+r^2+81-18r+r^2
r^2-34r-145=0
(r-29)(r-5)=0
r=29, or r=5
check both answers.
then put it in standard form.
Determine an equation of the circle that is tangent to both axes, has its center in the fourth quadrant and passes through the point (8, -9).
Help please?
Thanks much!
1 answer
2 answers