Question #138699
What is the equation of the circle with center at (2,5) passing through (-5,5) and (-1,1)
1
Expert's answer
2020-10-20T17:53:50-0400

The equation of the circle of radius RR with center at (a,b)(a,b) is (xa)2+(yb)2=R2(x-a)^2+(y-b)^2=R^2. In our case, (x2)2+(y5)2=R2(x-2)^2+(y-5)^2=R^2. Since the circle passing through (5,5)(-5,5), we have (52)2+(55)2=R2(-5-2)^2+(5-5)^2=R^2, and consequently R=7R=7. On the other hand, since the circle passing through (1,1)(-1,1), we have (12)2+(15)2=R2(-1-2)^2+(1-5)^2=R^2, and therefore R=5R=5. This contradiction shows that such a circle does not exist.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS