Question 986754
{{{x^2-2x+y^2-3y-1=0}}}
{{{(-1)^2-2(-1)+(2)^2-3(2)-1=0}}}
{{{1+2+4-6-1=0}}}
{{{0=0}}}
True, so point E satisfies the equation and therefore lies on the circle.
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Complete the square in x and y to find the center of the circle,{{{x^2-2x+y^2-3y-1=0}}}
{{{(x^2-2x+1)+(y^2-3y+9/4)-1=1+9/4}}}
{{{(x-1)^2+(y-3/2)^2=2+9/4}}}
{{{(x-1)^2+(y-3/2)^2=17/4}}}
So the center of the circle is ({{{1}}},{{{3/2}}})
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The x distance from E to the center is the same as the x distance from the center to F.
{{{d[x]=1-(-1)=2}}}
{{{F[x]=1+2=3}}}
Similarly for y,
{{{d[y]=3/2-2=-1/2}}}
{{{F[y]=3/2-1/2=1}}}
So point F is ({{{3}}},{{{1}}})
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*[illustration fd12.JPG].