What conic section has the equation $x^{2} + y^{2} + 12 x + 8 y = 48$ $x^2+y^2+12x+8y=48$ ?

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What conic section has the equation $x^{2} + y^{2} + 12 x + 8 y = 48$ $x^2+y^2+12x+8y=48$ ?

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Answer 1 · 2014-09-13T00:03:20

This is an equation for a circle. You begin by reorganizing the terms of the function so that $x$ $x$ and $x^{2}$ $x^2$ are together and $y$ $y$ and $y^{2}$ $y^2$ are together.

Next you will have to use the Completing the Square method.

Step 1: Reorder the terms

$x^{2} + 12 x + y^{2} + 8 y = 48$ $x^2+12x+y^2+8y=48$

Step 2: Begin Completing the square

$x^{2} + 12 x + y^{2} + 8 y = 48$ $x^2+12x+y^2+8y=48$

${(\frac{12}{2})}^{2} = 6^{2} = 36$ $(12/2)^2=6^2=36$ , Value to be added to complete the square

${(\frac{8}{2})}^{2} = 4^{2} = 16$ $(8/2)^2=4^2=16$ , Value to be added to complete the square

$x^{2} + 12 x + 36 + y^{2} + 8 y + 16 = 48 + 36 + 16$ $x^2+12x+36+y^2+8y+16=48+36+16$

$(x^{2} + 12 x + 36) + (y^{2} + 8 y + 16) = 100$ $(x^2+12x+36)+(y^2+8y+16)=100$

Factor

${(x + 6)}^{2} + {(y + 4)}^{2} = 100$ $(x+6)^2+(y+4)^2=100$

Solution: Standard form of a Circle.

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What conic section has the equation $x^{2} + y^{2} + 12 x + 8 y = 48$ $x^2+y^2+12x+8y=48$ ?

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What conic section has the equation $x^{2} + y^{2} + 12 x + 8 y = 48$ $x^2+y^2+12x+8y=48$ ?