Question 36547
TIP:
1.IF X^2 AND Y^2 HAVE EQUAL COEFFICIENTS AND THERE IS NO XY TERM ,IT
COULD BE A CIRCLE.
1.) X2 + y2 + 4x – 6y = 3
COMPLETE SQUARE.
(X^2+2*2X+2^2)-2^2+(Y^2+2*3Y+3^2)-3^2=3
(X+2)^2+(Y+3)^2=3+9+4=16=4^2
THIS IS THE EQN OF A CIRCLE WHOSE STD.FORM IS
(X-H)^2+(Y-K)^2=R^2,WHERE (H,K) IS THE CENTRE OF THE CIRCLE AND R IS
THE RADIUS.HENCE
(-2,-3) IS THE CENTRE AND 4 IS THE RADIUS.
{{{ graph( 500, 500, -10, 10, -10, 10,-3-(16-(x+2)^2)^0.5,-3+(16-(x+2)^2)^0.5  ) }}}