Find an equation for the circle centered at (4,-3) having radius at 2?

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Find an equation for the circle centered at (4,-3) having radius at 2?

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Answer 1 · 2018-05-01T21:00:50

$(x - 4) + (y + 3) = 4$ $(x-4)+(y+3)=4$

Explanation:

The standard equation of a circle is ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2+(y-b)^2=r^2$ where $(a, b)$ $(a,b)$ is the center and $r$ $r$ is the radius. Note the negatives for $a$ $a$ and $b$ $b$
See link for more detail https://www.khanacademy.org/math/algebra2/intro-to-conics-alg2/standard-equation-circle-alg2/v/writing-standard-equation-of-circle

Therefore, the equation of the circle with center $(4, - 3)$ $(4,-3)$ and radius $2$ $2$ will be that:
$a = 4$ $a=4$
$b = - 3$ $b=-3$
$r = 2$ $r=2$
Thus, the equation is ${(x - (4))}^{2} + {(y - (- 3))}^{2} = {(2)}^{2}$ $(x-(4))^2+(y-(-3))^2=(2)^2$ or simply:
${(x - 4)}^{2} + {(y + 3)}^{2} = 4$ $(x-4)^2+(y+3)^2=4$
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Find an equation for the circle centered at (4,-3) having radius at 2?

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