The circle (x-h)^2+(y-3)^2=36 passes through the origins. Find the values of h. Would you substitute 0,0 for x and y?
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Originally Posted by requal The circle (x-h)^2+(y-3)^2=36 passes through the origins. Find the values of h. Would you substitute 0,0 for x and y? Yes, that would be the first step.
yes, but that means I would get h^2+9=36 which becomes h^2=25 which therefore means that h=5,-5. However the answers says that it is
Originally Posted by requal yes, but that means I would get h^2+9=36 which becomes h^2=25 which therefore means that h=5,-5. However the answers says that it is In my universe 36 - 9 = 27 (most of the time, anyway).
eh, thanks anyway
Originally Posted by requal eh, thanks anyway I assume that means you understand where the correct answer has come from ....
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