Question 1088136
Use the general equation of a circle of radius R centered at (h,k),
{{{(x-h)^2+(y-k)^2=R^2}}}
{{{(x+3)^2+(y-6)^2=R^2}}}
Since it's tangent to the y-axis, (0,6) is also on the circle,
{{{(0+3)^2+(6-6)^2=R^2}}}
{{{R^2=9}}}
So,
{{{(x+3)^2+(y-6)^2=9}}}
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