How to find the coordinates of the center of the circle when equation is given and the equation is 2x^2 + 2y^2 - x = 0?

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Answer 1 · 2018-04-12T04:11:12

center #=(1/4,0)#

The coordinates center of circle with equation #(x-h)^2+(y-h)^2=r^2# is #(h,k)# where #r# is the radius of thee circle.

Given that,

#rarr2x^2+2y^2-x=0#

#rarr2(x^2+y^2-x/2)=0#

#rarrx^2-2*x*1/4+(1/4)^2-(1/4)^2+y^2=0#

#rarr(x-1/4)^2+(y-0)^2=(1/4)^2#

Comparing this with #(x-h)^2+(y-h)^2=r^2#, we get

#rarrh=1/4, k=0, r=1/4#

#rarr#center#=(h,k)=(1/4,0)#