Question

centro en el punto (-6, 1) y radio √5​

Answer 1

Respuesta:

${x}^{2} + {y}^{2} + 12x - 2y + 32 = 0$

Explicación paso a paso:

${(x - h)}^{2} + {(y - k)}^{2} = {( \sqrt{5} })^{2}$

${(x - ( - 6))}^{2} + {(y - 1)}^{2} = {( \sqrt{5})}^{2}$

${(x + 6)}^{2} + {(y - 1)}^{2} = 5$

${x}^{2} + 12x + 36 + {y}^{2} - 2y + 1 = 5$

${x}^{2} + {y}^{2} + 12x - 2y + 27 - 5 = 0$

${x}^{2} + {y}^{2} + 12x - 2y + 32 = 0$