ANSWERS: 3

If we take a circle on a graph, having the center on (0,0),we get the equation from Pythagorus theorem that (x0)^2 +(y0)^2 = a^2 Now, if we've to move it to some other coordinates the equation'd be (xA)^2 + (yB)^2 =a^2[where, (A,B) is center, a is radius] => x^2 + y^2 2Ax 2By +A^2 +B^2 a^2 =0 We can say now, x^2 + y^2 +2gx + 2fy + c =0..................(1) THIS IS THE STANDARD EQUATION OF A CIRCLE [If we substitute g = A, f = B, c = A^2 + B^2 a^2] [& a = root( A^2 + B^2 c)] Now, the coordinates of the center, (A,B)=(g,f) radius a = root( g^2 +f^2 c) [ replacing the values] *************************************************************************************** Here, now we move to your eqn & rewrite it x^2 + y^2 12x+6y=19 => x^2 + y^2 +2(6)x +2*3y +(19) =0.................(2) comparing (1) & (2); we get, center (A,B) = (g,f) =(6,3) radius a = root[ (6)^2 +3^2 (19)] =root(64) = 8 ***************************************************************************************** I think you've got your answer. If still you have the problem with it, do please write me a comment without hesitation.

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x2+y212x+6y=19 you may solve this by completing the square (x2 12x ____) +(y2 +6y ___) = 19 (x212x+36)+(y2+6y+9) =19 + 36 + 9 (x6)^2 (y+3)^2 = 64 therefore r = 8 (square root of 64) x = 6 y =3 location of the center
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