Question 1015646
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A).
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It is difficult to discern your equation.
If this is not your equation, clarify it and re-post your question.
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{{{(x-1)^2+(y-3)^2=0}}}
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Point (-1,1)
x=-1, y=1
Put these values in the equation, and evaluate to see if it is true.
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{{{((-1)-1)^2+(1-3)^2=0}}}
{{{-2)^2+(-2)^2=0}}}
{{{4+4=0}}}
{{{8=0}}} . False. The point is not on the graph.
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B).
x intercept: y=0
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{{{(x-1)^2+(y-3)^2=0}}}
{{{(x-1)^2+(0-3)^2=0}}}
{{{(x-1)^2+(-3)^2=0}}}
{{{(x-1)^2+9=0}}}
{{{(x-1)^2=-9}}}
{{{sqrt((x-1)^2)=sqrt(-9)}}}
{{{x-1}}}=+/-{{{3i}}}
{{{x=1+3i}}} {{{OR}}} {{{x=1-3i}}}
No x intercept.
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y intercept: x=0
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{{{(x-1)^2+(y-3)^2=0}}}
{{{(0-1)^2+(y-3)^2=0}}}
{{{(-1)^2+(y-3)^2=0}}}
{{{1+(y-3)^2=0}}}
{{{y-3)^2=-1}}}
{{{sqrt((y-3)^2)=sqrt(-1)}}}
{{{y-3}}}=+/-{{{i}}}
{{{y=3+i}}} {{{OR}}} {{{y=3-1}}}
No y intercept.
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Graph of the equation is a circle with center at (1,3) and zero radius.
The only real number solution is point (1,3).