Question 616135
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Hi, 
C: (0,0) and the point (-4,6) lies on the circle
D = {{{sqrt ((x[1]-x[2])^2+(y[1]-y[2])^2))}}}, r ={{{sqrt(4^2 + (-6)^2) = sqrt(52)}}} 
Standard Form of an Equation of a Circle is {{{(x-h)^2 + (y-k)^2 = r^2}}}
where Pt(h,k) is the center and r is the radius
 equation for this circle, in standard form: {{{x^2 + y^2 = 52}}}