How do you find the equation of the circle given center (-1,4) and radius 3?

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Answer 1 · 2016-10-30T10:09:22

#x^2+y^2+2x-8y+8=0#

If the center of circle is #(h,k)# and radius is #r#, its equation is

#(x-h)^2+(y-k)^2=r^2#

Hence equation of circle with center at #(-1,4)# and radius #3#, is

#(x-(-1))^2+(y-4)^2=3^2#

or #(x+1)^2+(y-4)^2=3^2#

or #x^2+2x+1+y^2-8y+16=9#

or #x^2+y^2+2x-8y+1+16-9=0#

or #x^2+y^2+2x-8y+8=0#