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

$x^{2} + y^{2} + 2 x - 8 y + 8 = 0$ $x^2+y^2+2x-8y+8=0$

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

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$

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

${(x - (- 1))}^{2} + {(y - 4)}^{2} = 3^{2}$ $(x-(-1))^2+(y-4)^2=3^2$

or ${(x + 1)}^{2} + {(y - 4)}^{2} = 3^{2}$ $(x+1)^2+(y-4)^2=3^2$

or $x^{2} + 2 x + 1 + y^{2} - 8 y + 16 = 9$ $x^2+2x+1+y^2-8y+16=9$

or $x^{2} + y^{2} + 2 x - 8 y + 1 + 16 - 9 = 0$ $x^2+y^2+2x-8y+1+16-9=0$

or $x^{2} + y^{2} + 2 x - 8 y + 8 = 0$ $x^2+y^2+2x-8y+8=0$