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# need help with coordinates

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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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Hello Guest,

$$x^2+y^2=4x-12y+8$$

$$x^2+y^2-4x+12y=8$$

$$x^2-4x+y^2+12y=8$$

$$x^2-4x+?+y^2+12y=8+?$$

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

$$x^2-4x+4+y^2+12y=8+4$$

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

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

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

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

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

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

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

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

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

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

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

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

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

Oct 29, 2021