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Find the area of the region enclosed by the graph of x^2 + y^2 = 4x - 12y + 8.

 Oct 28, 2021
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\(x^2+y^2=4x-12y+8\)

\(x^2+y^2-4x+12y=8\)

\(x^2-4x+y^2+12y=8\)

\(x^2-4x+?+y^2+12y=8+?\)

\(\mbox {To complete the square } {x}^{2}-4x+4=(x-2)^2 \mbox { add four}\)

\(x^2-4x+4+y^2+12y=8+4\)

\(\mbox {Use } a^2-2ab+b^2=(a-b)^2 \mbox { to factor the expression}\)

\((x-2)^2+y^2+12y=8+4\)

\((x-2)^2+y^2+12y=12\)

\(\mbox {To complete the square, the same value needs to be added to both sides}\)

\((x-2)^2+y^2+12y+?=12+?\)

\(\mbox {To complete the square } y^2+12y+36=(y+6)^2 \mbox { add 36 to the expression}\)

\((x-2)^2+y^2+12y+36=12+?\)

\(\mbox {Since 36 was added to the left side, also add 36 to the right side}\)

\((x-2)^2+y^2+12y+36=12+36\)

\(\mbox {Use the first binomial } a^2+2ab+b^2=(a+b)^2 \mbox { to factor the expression}\)

\((x-2)^2+(y+6)^2=12+36\)

\((x-2)^2+(y+6)^2=48\)

\(\mbox {The equation can be written in the form } (x-p)^2+(y-q)^2=r^2 \mbox { , so it represents a circle with the radius } r=\sqrt{48} \mbox { and the center } (2,-6)\)

 Oct 29, 2021

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