Question

X squared + ysquared-6y=-5

Answer 1

Standard Form of the circle:  

$x^{2} +(y-3) ^{2} =14$

For y:

y=3 (+/-) $\sqrt{-x} ^{2} +14$

For x:

x=$\sqrt{-y ^{2} } +6y+5 and - \sqrt{-y} ^{2} +6y+5$