ANSWERS: 3
  • If we take a circle on a graph, having the center on (0,0),we get the equation from Pythagorus theorem that (x-0)^2 +(y-0)^2 = a^2 Now, if we've to move it to some other co-ordinates the equation'd be (x-A)^2 + (y-B)^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 co-ordinates 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 & re-write 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.
  • See images. Let me know if you have any question or need anymore help.
  • x2+y2-12x+6y=19 you may solve this by completing the square (x2 -12x ____) +(y2 +6y ___) = 19 (x2-12x+36)+(y2+6y+9) =19 + 36 + 9 (x-6)^2 (y+3)^2 = 64 therefore r = 8 (square root of 64) x = 6 y =-3 location of the center

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