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Difficulty: Hard

Circle A has equation ${(x - 7)}^{2} + {(y + 3)}^{2} = 1$ . In the $x y$ -plane, circle B is obtained by translating circle A to the right $4$ units. Which equation represents circle B?

${(x - 7)}^{2} + {(y + 7)}^{2} = 1$

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

${(x - 11)}^{2} + {(y + 3)}^{2} = 1$

${(x - 7)}^{2} + {(y - 1)}^{2} = 1$

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Explanation

Choice C is correct. The equation of a circle in the xy-plane can be written as ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ , where the center of the circle is $(h, k)$ and the radius of the circle is $r$ units. It’s given that circle A has the equation ${(x - 7)}^{2} + {(y + 3)}^{2} = 1$ , which can be written as ${(x - 7)}^{2} + (y - {(- 3))}^{2} = 1^{2}$ . It follows that $h equals 7$ , $k = - 3$ , and $r equals 1$ . Therefore, the center of circle A is $left parenthesis 7 comma negative 3 right parenthesis$ and its radius is $1$ unit. If circle A is translated $4$ units to the right, the x-coordinate of the center will increase by $4$ , while the y-coordinate and the radius of the circle will remain unchanged. Translating the center of circle A to the right $4$ units yields $(7 + 4, - 3)$ , or $left parenthesis 11 comma negative 3 right parenthesis$ . Therefore, the center of circle B is $left parenthesis 11 comma negative 3 right parenthesis$ . Substituting $11$ for $h$ , $negative 3$ for $k$ , and $1$ for $r$ into the equation ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ yields ${(x - 11)}^{2} + (y - {(- 3))}^{2} = 1$ , or ${(x - 11)}^{2} + {(y + 3)}^{2} = 1$ . Therefore, the equation ${(x - 11)}^{2} + {(y + 3)}^{2} = 1$ represents circle B.

Choice A is incorrect. This equation represents a circle obtained by shifting circle A down, rather than right, $4$ units.

Choice B is incorrect. This equation represents a circle obtained by shifting circle A left, rather than right, $4$ units.

Choice D is incorrect. This equation represents a circle obtained by shifting circle A up, rather than right, $4$ units.