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A certain circle can be represented by the following equation. x^2+y^2+16x-14y+49=0x 2 +y 2 +16x−14y+49=0x, squared, plus, y, squared, plus, 16, x, minus, 14, y, plus, 49, equals, 0 What is the center of this circle ? ((left parenthesis ,,comma ))right parenthesis What is the radius of this circle ? units

Respuesta :

semsee45

Answer:

center = (-8, 7)

radius = 8 units

Step-by-step explanation:

Equation of a circle: [tex](x-h)^2+(y-k)^2=r^2[/tex]

(where (h, k) is the center of the circle and r is the radius)

Rewrite the given equation:

[tex]x^2+y^2+16x-14y+49=0[/tex]

[tex]\implies x^2+16x+y^2-14y+49=0[/tex]

[tex]\implies (x+8)^2-64+(y-7)^2-49+49=0[/tex]

[tex]\implies (x+8)^2+(y-7)^2=64[/tex]

Therefore:

  • center = (-8, 7)
  • radius:  [tex]r^2=64 \implies r=\sqrt{64} =8[/tex]
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