Asked by Sum3a
find the equation of a circle passing through the points p(0,-3) and q(4,0) and has its centre on the line x+2y=0
Answers
Answered by
Steve
The circle's equation is
(x-h)^2 + (y-k)^2 = r^2
You know that
(0-h)^2 + (-3-k)^2 = r^2
(4-h)^2 + (0-k)^2 = r^2
h+2k = 0
Now solve for h,k,r
I'd start with
h^2+(k+3)^2 = (h-4)^2+k^2
h = -2k, so
4k^2 + (k+3)^2 = (-2k-4)^2 + k^2
solve that for k, and then you can get h and r.
(x-h)^2 + (y-k)^2 = r^2
You know that
(0-h)^2 + (-3-k)^2 = r^2
(4-h)^2 + (0-k)^2 = r^2
h+2k = 0
Now solve for h,k,r
I'd start with
h^2+(k+3)^2 = (h-4)^2+k^2
h = -2k, so
4k^2 + (k+3)^2 = (-2k-4)^2 + k^2
solve that for k, and then you can get h and r.
Answered by
Sum3a
Thanks a lot steve .... appreciate that !!
But can i use elimination method to solve for h and k ?
But can i use elimination method to solve for h and k ?
Answered by
Steve
sure - do whatever works for you.
Answered by
Anonymous
Thanks for ur help
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