Question

por favor ayúdame

plissss
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Answer 1

Respuesta:

Espero sea completar la ecuacion de la circunferencia.

El radio de cada literal sería

$c) \: r = 7 \: \: \: \: \: \: d) \: r = \sqrt{5} \: \: \: \: \: \: e) \: r = 4$

Explicación paso a paso:

$c) {x}^{2} + {y}^{2} + 2x - 48 = 0 \\ ( {x}^{2} + 2x) + {y}^{2} = 48 \\ ( {x}^{2} + 2x + 1 - 1) + {y}^{2} = 48 \\ ( {(x + 1)}^{2} - 1 )+ {y}^{2} = 48 \\ {(x + 1)}^{2} + {y}^{2} = 48 +1 \\ {(x + 1)}^{2} + {y}^{2} = 49 \\ {(x + 1)}^{2} + {y}^{2} = {(7)}^{2}$

$d) {x}^{2} + {y}^{2} - 2x - 4y = 0 \\ ( {x}^{2} - 2x) + ( {y}^{2} - 4y) = 0 \\ ( {x}^{2} - 2x + 1 - 1) + ( {y}^{2} - 4y + 4 - 4) = 0 \\ ( {(x - 1)}^{2} - 1) + ( {(y - 2)}^{2} - 4) = 0 \\ {(x - 1)}^{2} + {(y - 2)}^{2} = 1 + 4 \\ {(x - 1)}^{2} + {(y - 2)}^{2} = 5\\{(x - 1)}^{2} + {(y - 2)}^{2} = {( \sqrt{5} )}^{2}$

$e) {x}^{2} + {y}^{2} - 8x = 0 \\ ( {x}^{2} - 8x) + {y}^{2} = 0 \\ ( {x}^{2} - 8x + 16 - 16) + {y}^{2} = 0 \\ ( {(x - 4)}^{2} - 16) + {y}^{2} = 0 \\ {(x - 4)}^{2} + {y}^{2} = 16 \\ {(x - 4)}^{2} + {y}^{2} = {(4)}^{2}$