Question

Obtenha o centro e o raio da circunferência da equação 2x²+2y²-4x-6y-6=0

Answer 1

Olá!

Completemos o quadrado!

$\\ \displaystyle \mathsf{2x^2 + 2y^2 - 4x - 6y - 6 = 0} \\\\ \mathsf{2(x^2 - 2x + y^2 - 3y - 3) = 0} \\\\ \mathsf{x^2 - 2x + y^2 - 3y - 3 = 0} \\\\ \mathsf{(x^2 - 2x) + (y^2 - 3y) = 3} \\\\ \mathsf{[(x^2 - 2x + 1) - 1] + \left [ \left ( y^2 - 3y + \frac{9}{4} \right ) - \frac{9}{4}\right ] = 3} \\\\ \mathsf{(x - 1)^2 - 1 + \left ( y - \frac{3}{2} \right )^2 - \frac{9}{4} = 3}$

$\\ \displaystyle \mathsf{(x - 1)^2 + \left ( y - \frac{3}{2} \right )^2 = \frac{9}{4} + 1 + 3} \\\\\\ \mathsf{(x - 1)^2 + \left ( y - \frac{3}{2} \right )^2 = \frac{9}{4} + 4} \\\\\\ \mathsf{(x - 1)^2 + \left ( y - \frac{3}{2} \right )^2 = \frac{25}{4}} \\\\ \boxed{\mathsf{(x - 1)^2 + \left ( y - \frac{3}{2} \right )^2 = \left ( \frac{5}{2} \right )^2}}$

 Assim, temos que o centro é (1, 3/2) e o raio 5/2.