Question 866674
Using the standard form equation for a circle, you can make three separate equations for a system:


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{{{(7-h)^2+(6-k)^2=r^2}}}
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{{{(11-h)^2+(6-k)^2=r^2}}}
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{{{(9-h)^2+(4-k)^2=r^2}}}
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Equating the left members of the first two equations, based on both equal to {{{r^2}}} and that both members contain {{{(6-k)^2}}}, allows you to find a value for h.


You can use this value for h to turn the variable expression in each of the three equations into a constant term; so now you will have three equations and just TWO unknowns, k and r.


I started this separately myself on paper, but not yet finished...
This part just for h, but omitting to show the steps of the process:  {{{highlight_green(h=-2)}}}...


Using that {{{h=-2}}}, the equation Number 2 and Number 3 become:
{{{169+(6-k)^2=r^2}}}
and
{{{121+(4-k)^2=r^2}}}
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Now equation each expression for both being {{{r^2}}}, this gives {{{169+(6-k)^2=121+(4-k)^2}}}.
Performing the arithmetic, algebraic steps gives...
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{{{highlight_green(k=17)}}}.


Use any of the original or transformed equations of the system to compute the value of r.  
{{{81+(6-(-17))^2=r^2}}}
{{{81+23^2=r^2}}}
{{{r^2=81+529}}}
{{{highlight_green(r^2=610)}}}-----see that this is still in the squared form.
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The equation of the circle should be:  {{{highlight((x+2)^2+(y-17)^2=610)}}}.