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Question; Decide whether of not the equations has a circle as its graph. If it does, give the center and the radius. #20. x^2-12x+y^2+20=0 Thank you so much for your help. Angela
\n" ); document.write( "THERE ARE 4 CRITERIA FOR A GENERAL SECOND DEGREE EQN. IN X AND Y TO REPRESENT A CIRCLE..
\n" ); document.write( "1.EQN.SHOULD BE SECOND DEGREE IN X AND Y.....OK
\n" ); document.write( "2.COEFFICIENTS OF X^2 AND Y^2 SHOULD BE EQUAL...OK...HERE BOTH ARE 1
\n" ); document.write( "3.THERE SHOULD BE NO XY TERM.......OK.........IT IS NOT THERE
\n" ); document.write( "4.THIS INVOLVES A RELATION BETWEN COEFFICIENTS OF VARIOUS TERMS...BUT LET ME PUT IT IN SIMPLE TERMS TO SAY THAT WHEN THE ABOVE EQUATION IS WRITTEN IN THE STADARD FORM AS BELOW....THE R.H.S.SHALL BE A POSITIVE NUMBER
\n" ); document.write( "(X-H)^2+(Y-K)^2=R^2....WHERE (H,K) IS THE CENTRE OF THE CIRCLE AND R IS IT RADIUS..
\n" ); document.write( "THE ABOVE EQN. CAN BE PUT AS
\n" ); document.write( "(X^2-2*X*6+6^2)+(Y^2)+20-6^2=0...OR...
\n" ); document.write( "(X-6)^2+(Y-0)^2=16...WHOSE RHS IS A POSITIVE NUMBER....SO THIS IS A CIRCLE AND COMPARING WITH ABOVE STANDARD EQN.WE CAN CONCLUDE THAT ITS CENTRE IS (6,0) AND RADIUS=4
\n" ); document.write( "OK...THE GENERAL RELATION BETWEEN COEFFICIENTS IS
\n" ); document.write( "G^2+F^2-C=POSITIVE..WHERE G IS HALF THE COEFFICIENT OF X....=-6 HERE
\n" ); document.write( "F IS HALF THE COEFFICIENT OF Y....=0 HERE
\n" ); document.write( "C IS THE CONSTANT TERM ....=20 HERE
\n" ); document.write( "(-6)^2+0^2-20=36-20=16 POSITIVE....OK.... \n" ); document.write( "