Question

Which of the following equations are equations of circles? Choose all that apply.

Answer 1

Number 3 and Number 4

Explanation:

Answer 2

`x^(2)+y^(2)+2x+10y-5=0`

`3x^(2)+3y^(2)+12x+30y-6=0`

Explanation:

we know that

The equation of a circle ion standard form is equal to

`(x-h)^(2) +(y-k)^(2)=r^(2)`

where

(h,k) is the center and r is the radius

Verify each case

case 1) we have

`(x+5)^(2) +(y+2)^(2)=-4`

The radius squared cannot be a negative number

`r^(2)\\eq -4`

therefore

This equation is not the equation of a circle

case 2) we have

`2x^(2)+y^(2)+15x-y+3=0`

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`(2x^(2)+15x)+(y^(2)-y)=-3`

Factor the leading coefficient of each expression

`2(x^(2)+7.5x)+(y^(2)-y)=-3`

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

`2(x^(2)+7.5x+14.0625)+(y^(2)-y+0.25)=-3+28.125+0.25`

`2(x^(2)+7.5x+14.0625)+(y^(2)-y+0.25)=25.375`

Rewrite as perfect squares

`2(x+3.75)^(2)+(y-0.5)^(2)=25.375`

This equation is not the equation of a circle

case 3) we have

`x^(2)+y^(2)+2x+10y-5=0`

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`(x^(2)+2x)+(y^(2)+10y)=5`

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

`(x^(2)+2x+1)+(y^(2)+10y+25)=5+1+25`

`(x^(2)+2x+1)+(y^(2)+10y+25)=31`

Rewrite as perfect squares

`(x+1)^(2)+(y+5)^(2)=31`

This equation represent the equation of a circle

case 4) we have

`3x^(2)+3y^(2)+12x+30y-6=0`

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`(3x^(2)+12x)+(3y^(2)+30y)=6`

Factor the leading coefficient of each expression

`3(x^(2)+4x)+3(y^(2)+10y)=6`

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

`3(x^(2)+4x+4)+3(y^(2)+10y+25)=6+12+75`

`3(x^(2)+4x+4)+3(y^(2)+10y+25)=93`

Rewrite as perfect squares

`3(x+2)^(2)+3(y+5)^(2)=93`

Divide by 3 both sides

`(x+2)^(2)+(y+5)^(2)=31`

This equation represent the equation of a circle

case 5) we have

`x^(2)-y^(2)+20x+2y+9=0`

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`(x^(2)+20x)-(y^(2)-2y)=-9`

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

`(x^(2)+20x+100)-(y^(2)-2y+1)=-9+10-1`

`(x^(2)+20x+100)-(y^(2)-2y+1)=0`

Rewrite as perfect squares

`(x+10)^(2)-(y-1)^(2)=0`

This equation is not the equation of a circle