Question 957021
{{{(x-3)^2 =9- (y+3)^2}}}
{{{x-3=0 +- sqrt(9- (y+3)^2)}}}
{{{x=3 +- sqrt(9- (y+3)^2)}}}
So then
{{{x^2=9+6sqrt(9-(y+3)^2)+9-(y+3)^2}}}-Positive
{{{x^2=9-6sqrt(9-(y+3)^2)+9-(y+3)^2}}}-Negative
Substituting,
{{{(9+6sqrt(9-(y+3)^2)+9-(y+3)^2) + y^2 = 9}}}-Positive
{{{(9-6sqrt(9-(y+3)^2)+9-(y+3)^2) + y^2 = 9}}}-Negative
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{{{6sqrt(9-(y+3)^2)+9-y^2-6y-9 + y^2 = 0}}}-Positive
{{{-6sqrt(9-(y+3)^2)+9-y^2-6y-9 + y^2 = 0}}}-Negative
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{{{sqrt(9-(y+3)^2)=y  }}}-Positive
{{{-sqrt(9-(y+3)^2)=y  }}}-Negative
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{{{9-(y+3)^2=y^2}}}
{{{9-y^2-6y-9=y^2}}}
{{{2y^2+6y=0}}}
{{{y(y+3)=0}}}
Two solutions:
{{{y=0}}}
Then,
{{{(x-3)^2 + (0+3)^2 =9}}}
{{{(x-3)^2=0}}}
{{{x=3}}}
and
{{{y+3=0}}}
{{{y=-3}}}
Then,
{{{x^2+(-3)^2=9}}}
{{{x=0}}}
(3,0) and (0,-3)
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{{{drawing(300,300,-4,6,-6,4,grid(1),circle(0,0,3),circle(3,-3,3),circle(3,0,0.2),circle(0,-3,0.2))}}}