Asked by Nikki
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!
Help please?
Thanks much!
Answers
Answered by
bobpursley
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.