Question

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used. 

Answer 1

The standard form of the equation of a circumference is given by the following expression:

$(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$

On the other hand, the general form is given as follows:

$x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

$\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

First. a) matches 5)

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

Second. b) matches 1)

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

Third. c) matches 3)

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

Fourth. d) matches 6)

Answer 2

If the General equation of circle is

$x^2+y^2+2 g x + 2 f y+c=0\\\\ (x+g)^2+(y+f)^2=\sqrt{(g^2+f^2-c)^2$

$1.x^2 + y^2 + 4 x + 12 y - 20 = 0 \\\\ (x+2)^2+(y+6)^2=60\\\\ 2. x^2 + y^2 + 6 x - 8 y -10 = 0\\\\ (x+3)^2+(y-4)^2 =35\\\\3. 3x^2 + 3y^2 + 12 x + 18 y - 15 = 0\\\\ x^2 +y^2+4 x+ 6 y-5=0\\\\ (x+2)^2+(y+3)^2=18\\\\4. 5x^2 + 5y^2 - 10 x + 20y -30 = 0\\\\ x^2+y^2-2 x + 4 y -6=0\\\\ (x-1)^2+(y+2)^2=11\\\\5. 2x^2 + 2y^2 - 24x - 16y - 8 = 0\\\\ x^2 +y^2-12 x-8 y-4=0\\\\ (x-6)^2+(y-4)^2=56\\\\6. x^2 + y^2 + 2 x - 12 y - 9 = 0 \\\\ (x+1)^2+(y-6)^2=46$