How do you find the equation given center passes through the origin and has its center at (0,4)?

1 Answer
Alan P.
Dec 23, 2016

#x^2+(y-4)^2=16#

Explanation:

Assuming this is to be the equation of a circle (although it was asked under the more general "Geometry of an Ellipse")

A circle with center #(color(red)a,color(blue)b)# and radius #color(green)r# has the equation:
#color(white)("XXX")(x-color(red)a)^2+(y-color(blue)b)^2=color(green)r^2#

If the required circle has its center at #(0,4)#
then #color(red)a=color(red)0#
and #color(blue)b=color(blue)4#

Further if its center is at #(0,4)# and it passes through the origin i.e. through #(0,0)# it radius must be #color(green)r=color(green)4#

So the equation of the required circle must be
#color(white)("XXX")(x-color(red)0)^2+(y-color(blue)4)^2=color(green)4^2#
or, simplifying slightly:
#color(white)("XXX")x^2+(y-4)^2=16#

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