How do you write the standard equation of a circle with the given radius and center: r=4; C(3,-4)?

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Answer 1 · 2018-05-17T16:18:03

See below

If you know the center $(x_{0}, y_{0})$ $(x_0,y_0)$ and the radius $r$ $r$ of a circle, you can write its equation as

${(x - x_{0})}_{0}^{2} + {(y - y_{0})}_{0}^{2} = r^{2}$ $(x-x_0)^2+(y-y_0)^2=r^2$

which in your case is

${(x - 3)}^{2} + {(y + 4)}^{2} = 16$ $(x-3)^2+(y+4)^2=16$

you can expand it into

$x^{2} - 6 x + 9 + y^{2} + 8 y + 16 = 16$ $x^2-6x+9+y^2+8y+16=16$

and simplify/rearrange into

$x^{2} + y^{2} - 6 x + 8 y + 9 = 0$ $x^2+y^2-6x+8y+9=0$

Not sure which one you refer to as "standard form", so I hope that one of these calculations fell into it