X squared + ysquared-6y=-5

1 answer:

8 0

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$

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