I need to solve the radius and the center point. $$ x^2+y^2=6x-2y-8 $$ Under this is my attempt of solving it $$ x^2-6x+y^2+2y=-8 $$ $$ x^2-6x+9+y^2\ +2x+1=-8+1+9 $$ $$ \left(x+3\right)^2+\left(y-1\right)^2=2^2 $$ On 2nd I added the integers to both sides
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5The last formula has more than 2 errors – Lutz Lehmann Dec 01 '23 at 09:48
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Expand your last line out. – Paul Dec 01 '23 at 10:04
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1The 2nd to last line also has an error. – user2661923 Dec 01 '23 at 10:06
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what is your question? – Sine of the Time Dec 01 '23 at 14:40
3 Answers
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By solving you can get $$x^2+y^2=6x−2y−8$$
$$(x^2-6x+9-9) + (y^2+2y+1-1) =-8$$
$$(x-3)^2 +(y+1)^2 = 2.$$
Anne Bauval
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Antarjot
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1https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Sine of the Time Dec 01 '23 at 14:38
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Close but no cigars :-)
At the right side you have $=2$, which does not equal $=2^2$ but $=(\sqrt{2})^2$
:-)
Dominique
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$x^2+y^2=6x-2y-8$
$\Rightarrow x^2-6x+y^2+2y=-8$
$\Rightarrow x^2-6x+9+y^2+2y+1=-8+1+9$
(You have written wrongly $2x$ instead of $2y$)
$\Rightarrow (x^2-6x+9)+(y^2+2y+1)=2$
$\Rightarrow \left(x-3\right)^2+\left(y+1\right)^2={(\sqrt2)}^2$
(You have done wrong calculation here)
We know that,the equation of a circle with center at $(a,b)$ and with radius $r$ is, ${(x-a)}^2+ {(y-b)}^2= r^2$.
So comparing with above equation we get that, the center is $(3,-1)$ and radius is $\sqrt 2$.
Math solver
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