Question 385414
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Hi, 
Standard Form of an Equation of a Circle is {{{(x-h)^2 + (y-k)^2 = r^2}}}        
4x^2+4y^2+24x-56y+168 = 0 
4x^2+ 24x + 4y^2-56y+168 = 0 
x^2 +6x +y^2 - 14y + 42 = 0 Completing both squares
 (x+3)^2 -9 + (y -7)^2 -49 + 42 = 0
 (x+3)^2 (y -7)^2 -16 = 0
 (x+3)^2 (y -7)^2 = 16 
Center is Pt(-3,7)
{{{drawing(500,500, -15,3,-3,15,grid(1),circle(-3,7,4), circle(-3,7,.15), line(-3,7,1,7),locate(-3,7,C),locate(-1,7,r))}}}