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The equation of a given circle in general form is x2+ y2 − 8x + 12y + 27 = 0. Write the equation in standard form, (x − h)2 + (y - k)2 = r2, by completing the squares in the equation. Show your work in a table.

Respuesta :

The equation of the circle is [tex] x^{2} + y^{2} +8x+12y+27=0[/tex]

we take x-es and y-s 'close' to each other, and complete the square:

[tex]x^{2}+8x + y^{2} +12y+27=0[/tex]

write the coefficients of the linear terms (the x and y with degree 1) as 2*'something', to see how to complete the square:

[tex]x^{2}+2*4*x + y^{2} +2*6*y+27=0[/tex]

which means that [tex]x^{2}+2*4*x[/tex] needs [tex] 4^{2} [/tex]  

and [tex]y^{2} +2*6*y[/tex] needs [tex] 6^{2} [/tex] to become perfect square trinomials:

[tex](x^{2}+2*4*x + 4^{2})-4^{2} + (y^{2} +2*6*y+ 6^{2})-6^{2} +27=0[/tex]

[tex] (x+4)^{2}+ (y+6)^{2}=16+36-27 [/tex]
[tex] (x+4)^{2}+ (y+6)^{2}=25 [/tex]

[tex](x-(-4))^{2}+ (y-(-6))^{2}= 5^{2} [/tex]


Answer: [tex](x-(-4))^{2}+ (y-(-6))^{2}= 5^{2} [/tex]

cutelittlebean

Answer:

Step                                                                     Reason

x2 + y2 - 8x + 12y + 27 = 0                                 given

x2 - 8x + y2 + 12y = -27                                      Isolate the constant term

x2 - 8x + 16 + y2 + 12y + 36 = -27 + 16 + 36      Complete the square by adding                              .                                                                           16 and 36 to both sides.

(x2 - 2∙4∙x + 42) + y2 + 12y + 36 = 25                Group and rearrange terms in x.

(x - 4)2 + y2 + 12y + 36 = 25                              (a - b)2 = a2 - 2ab + b2

(x - 4)2 + (y2 + 2∙6∙y + 62) = 25                         Group and rearrange terms in y.

(x - 4)2 + (y + 6)2 = 25                                        (a + b)2 = a2 + 2ab + b2

(x - 4)2 + (y + 6)2 = 52                                        Take the square root to find r = 5.

Step-by-step explanation:

this is the exact answer from plato

hope this helps a little bit :))

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