kay10k kay10k
  • 22-05-2019
  • Mathematics
contestada

what is the center of the circle given by the equation x^2+y^2-14y-15=0

Respuesta :

gmany
gmany gmany
  • 22-05-2019

Answer:

(0, 7)

Step-by-step explanation:

The equation of a circle in the standard form:

[tex](a-h)^2+(y-k)^2=r^2[/tex]

(h, k) - center

r - radius

We have the equation:

[tex]x^2+y^2-12y-15=0[/tex]

Convert into a standard form using

[tex](a-b)^2=a^2-2ab+b^2\qquad(*)[/tex]

[tex]x^2+\underbrace{y^2-2(y)(7)+7^2}_{(*)}-7^2-15=0\\\\(x-0)^2+(y-7)^2-49-15=0[/tex]

[tex](x-0)^2+(y-7)^2-64=0[/tex]           add 64 to both sides

[tex](x-0)^2+(y-7)^2=64[/tex]

The center (0, 7)

The radius: r = √64 = 8

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