Question

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

x2 + y2 − 4x + 12y − 20 = 0
(x − 6)2 + (y − 4)2 = 56
x2 + y2 + 6x − 8y − 10 = 0
(x − 2)2 + (y + 6)2 = 60
3x2 + 3y2 + 12x + 18y − 15 = 0
(x + 2)2 + (y + 3)2 = 18
5x2 + 5y2 − 10x + 20y − 30 = 0
(x + 1)2 + (y − 6)2 = 46
2x2 + 2y2 − 24x − 16y − 8 = 0
x2 + y2 + 2x − 12y − 9 = 0

Answer 1

The standard forms of the given circle equations in general form are (x-2)^2 + (y+6)^2 = 60, (x+2)^2 + (y+3)^2 = 18, and (x+1)^2 + (y-6)^2 = 46.

Step-by-step explanation:

The standard form of a circle equation is (x-h)^2 + (y-k)^2 = r^2, where (h,k) represents the center of the circle and r represents the radius. To match the given equations in general form to their corresponding standard form equations:

x^2 + y^2 - 4x + 12y - 20 = 0 corresponds to (x-2)^2 + (y+6)^2 = 60

x^2 + y^2 + 6x - 8y - 10 = 0 corresponds to (x+2)^2 + (y+3)^2 = 18

5x^2 + 5y^2 - 10x + 20y - 30 = 0 corresponds to (x+1)^2 + (y-6)^2 = 46

Answer 2

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√(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√(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√(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=√(46) \\ D=2, \ E=-12, \ F=-9`

Then, its general form is:
`x^(2)+y^(2)+2x-12y-9=0`

Fourth. d) matches 6)