Determine the centre and radius of a circle given by the equation $2x^2\;+\;2y^2\;-\;12\;=\;4x\;-\;12y$

Determine the centre and radius of a circle given by the equation
$2x^2\;+\;2y^2\;-\;12\;=\;4x\;-\;12y$

Answers

$2x^2\; -\;4x\;+\;2y^2\;+12y\;-\;12$ = 0

Divide each term by 2

$x^2\;-2x +\;y^2 + 6y$ - 6 = 0

Complete the squares

(x - 1)² + (y + 3)² - 6 = 0

Centre (1,-3)

Radius = $\sqrt{6+(1)^2+(-3)^2}$

= $\sqrt{6+1+9}$ = $\sqrt16\$ = 4 units

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johnmulu answered the question on March 8, 2017 at 05:00

Next: Solve for y given that $log_2(2y\; + \;3)-\;2\; = log_2(y - 2)$
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