goatedvon goatedvon
  • 17-09-2019
  • Mathematics
contestada

What is the standard form of this equation of a circle?
x2 + y2 + 14x - 4y - 28 = 0

Respuesta :

DeanR
DeanR DeanR
  • 17-09-2019

We complete the square.

[tex]x^2 + y^2 + 14x - 4y - 28 = 0[/tex]

[tex]x^2+14x + y^2 - 4y = 28[/tex]

[tex]x^2+14x + (14/2)^2 + y^2 - 4y + (4/2)^2 = 28 + 49 + 4 = 81[/tex]

So the answer is:

[tex](x+7)^2 + (y-2)^2 = 9^2[/tex]

Answer Link
dora259
dora259 dora259
  • 01-07-2021

Answer:

(x + 7^)2 + (y − 2)^2 = 81

Step-by-step explanation:

(x2 + 14x + 49) + (y2 − 4y + 4) − 28 = 49 + 4

(x + 7)2 + (y − 2)2 − 28 = 49 + 4

(x + 7)2 + (y − 2)2 = 49 + 4 + 28 = 81.

Answer Link

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