The standard form of the equation of a circumference is given by the following expression:
[tex](x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius[/tex]
On the other hand, the general form is given as follows:
[tex]x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}[/tex]
In this way, we can order the mentioned equations as follows:
Equations in Standard Form:
[tex]\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46[/tex]
Equations in General Form:
[tex]\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0 [/tex]
[tex]\bold{6)} \ x^{2}+y^{2}+2x-12y[/tex]
So let's match each equation:
[tex]\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4[/tex]
Then, its general form is:
[tex]x^{2}+y^{2}-12x-8y-4=0[/tex]
First. a) matches 5)
[tex]\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20[/tex]
Then, its general form is:
[tex]x^{2}+y^{2}-4x+12y-20=0[/tex]
Second. b) matches 1)
[tex]\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5[/tex]
Then, its general form is:
[tex]x^{2}+y^{2}+4x+6y-5=0[/tex]
Third. c) matches 3)
[tex]\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9[/tex]
Then, its general form is:
[tex]x^{2}+y^{2}+2x-12y-9=0[/tex]
Fourth. d) matches 6)
