SOLUTION: Passes through the points of intersection of the circles x²+y²+2x-4x+6y=12, x²+y²+4x-8y=28 and has its center on the y axis.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Passes through the points of intersection of the circles x²+y²+2x-4x+6y=12, x²+y²+4x-8y=28 and has its center on the y axis.       Log On


   

Question 1143668: Passes through the points of intersection of the circles x²+y²+2x-4x+6y=12, x²+y²+4x-8y=28 and has its center on the y axis.

Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
What does?
There are many things that can pass thru 2 points and have a center on the y-axis.
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You want us to do the work to find the equation of a circle, but you don't have the time to say that?

Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.

This post has a  FATAL  DEFICIENCY : 

    (a)  it is UNCOMPLETED,

and

    (b)  it has no question.


It is just enough to re-direct it to the GARBAGE BOX.

And please, do not communicate with us as if you are a colonel or a general in a colonial occupying army.