Write the standard form of the equation of the circle with the center at (4,2) that has radius 4.A) ( x − 4 ) 2 + ( y − 2 ) 2 = 4 ( x - 4 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 4 ( x − 4 ) 2 + ( y − 2 ) 2 = 4 B) ( x + 4 ) 2 + ( y + 2 ) 2 = 16 ( x + 4 ) ^ { 2 } + ( y + 2 ) ^ { 2 } = 16 ( x + 4 ) 2 + ( y + 2 ) 2 = 16 C) ( x + 4 ) 2 + ( y + 2 ) 2 = 4 ( x + 4 ) ^ { 2 } + ( y + 2 ) ^ { 2 } = 4 ( x + 4 ) 2 + ( y + 2 ) 2 = 4 D) ( x − 4 ) 2 + ( y − 2 ) 2 = 16 ( x - 4 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 16 ( x − 4 ) 2 + ( y − 2 ) 2 = 16 E) ( x + 4 ) 2 + ( y − 2 ) 2 = 4 ( x + 4 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 4 ( x + 4 ) 2 + ( y − 2 ) 2 = 4

Question

Write the standard form of the equation of the circle with the center at (4,2)  that has radius 4.A)    ( x − 4 )  2 + ( y − 2 )  2 = 4 ( x - 4 )  ^ { 2 } + ( y - 2 )  ^ { 2 } = 4 ( x − 4 )  2 + ( y − 2 )  2 = 4 B)    ( x + 4 )  2 + ( y + 2 )  2 = 16 ( x + 4 )  ^ { 2 } + ( y + 2 )  ^ { 2 } = 16 ( x + 4 )  2 + ( y + 2 )  2 = 16 C)    ( x + 4 )  2 + ( y + 2 )  2 = 4 ( x + 4 )  ^ { 2 } + ( y + 2 )  ^ { 2 } = 4 ( x + 4 )  2 + ( y + 2 )  2 = 4 D)    ( x − 4 )  2 + ( y − 2 )  2 = 16 ( x - 4 )  ^ { 2 } + ( y - 2 )  ^ { 2 } = 16 ( x − 4 )  2 + ( y − 2 )  2 = 16 E)    ( x + 4 )  2 + ( y − 2 )  2 = 4 ( x + 4 )  ^ { 2 } + ( y - 2 )  ^ { 2 } = 4 ( x + 4 )  2 + ( y − 2 )  2 = 4

Jesse Rubin · Accepted Answer

Final Answer : D, Explanation :  The standard form of the equation of a circle with center $(h,k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$. Plugging in the values given, we get $(x-4)^2+(y-2)^2=4^2$, which simplifies to $(x-4)^2+(y-2)^2=16$. The correct choice is therefore D.

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