Question

Centro raio das seguintes circunferências:
X2+y2+2x-4y-44=0
X2+y2+2x-15=0

Answer 1

Resposta:

Formando o trinômio quadrado perfeito;

$x^{2}+2x+y^{2}-4y-44=0\\x^{2}+2x+y^{2}-4y=44\\(x+1)^{2}-1+(y-2)^{2}-4=44\\(x+1)^{2}+(y-2)^{2}=44+1+4\\(x+1)^{2}+(y-2)^{2}=49\\\\(x-x_{c})^{2}+(y-y_{c})^{2}=r^{2}\\r^{2}=49\\r=\sqrt{49}\\\boxed{raio=7}\\(x-(-1))^{2}+(y-2)^{2}=7^{2}\\\\\boxed{Centro}=(-1,2)\\\boxed{raio}=7$

$x^{2}+2x+y^{2}-15=0\\(x+1)^{2}-1+(y+0)^{2}-15=0\\(x+1)^{2}+y^{2}-1-15=0\\(x+1)^{2}+y^{2}=16\\\\r^{2}=16\\r=\sqrt{16}\\\boxed{raio=4}\\\\(x-x_{c})^{2}+(y-y_{c})^{2}=r^{2}\\(x+1)^{2}+(y-0)^{2}=16\\(x-(-1))^{2}+(y-0)^{2}=4^{2}\\\\\boxed{Centro}=(-1,0)\\\boxed{raio}=4$

Explicação passo-a-passo: