Question

Show that the equation represents a circle and find its center and radius. x2 + y2 +6y + 2 = 0

Answer 1

Answer:

Step-by-step explanation:

The general equation of a circle can be expressed as (x-a)²+(y-b)² = r² where (a,b) is the centre of the circle and r is the radius.

To show that the equation x² + y² +6y + 2 = 0 represents a circle, we need to write it in the format above using the completing the square method.

Given x² + (y² +6y) + 2 = 0

First we need to complete the square of the square in parenthesis by adding the square of half of the coefficient of y i.e $(\frac{1}{2}*6) ^2$ to the equation and adding the constant to the other side of the equation as well.

x² + (y² +6y) + 2 = 0

$x^2+(y^2+6y+(\frac{1}{2}*6) ^2)+2 = 0+(\frac{1}{2}*6) ^2\\x^2+(y^2+6y+(3) ^2)+2 = 0+3 ^2\\x^2+(y^2+6y+9)+2 = 0+9\\x^2+(y+3)^2+2 = 9\\x^2+(y+3)^2+2-2 = 9-2\\x^2+(y+3)^2 = 7\\(x-0)^2+(y+3)^2 = 7$

Hence the equation represents a circle with centre C at (0, -3) and radius of √7