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

1 Answer
Oct 30, 2016

x^2+y^2+2x-8y+8=0x2+y2+2x8y+8=0

Explanation:

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

(x-h)^2+(y-k)^2=r^2(xh)2+(yk)2=r2

Hence equation of circle with center at (-1,4)(1,4) and radius 33, is

(x-(-1))^2+(y-4)^2=3^2(x(1))2+(y4)2=32

or (x+1)^2+(y-4)^2=3^2(x+1)2+(y4)2=32

or x^2+2x+1+y^2-8y+16=9x2+2x+1+y28y+16=9

or x^2+y^2+2x-8y+1+16-9=0x2+y2+2x8y+1+169=0

or x^2+y^2+2x-8y+8=0x2+y2+2x8y+8=0