Find the radius and center of

Question

X² +y² - 4x+6y-3=0

Answer 1

To solve this, you need to complete the square, twice

Group the x and y terms together and add 3 to both sides of the equation.

The spaces are left for the c terms.

x²-4x+ y²+6y =3

The c terms are calculated with the formula (b/2)²

The x terms: (-4/2)² = 4 Add 4 to both sides

The y terms (6/2)² = 9 Add 9 to both sides

x²-4x+4 y²+6y+9 =3+4+9

Now on the left sides, factor both variables. Add the terms on the right side

(x-2)²+(y+3)² = 16

Now this is in standard form. The center would be (2,-3) and the radius is the square root of the term on the right. r=4

Answer 2

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

x^2+y^2-4x+6y-3=0

x^2-4x +y^2+6y =3

take half of -4 and square it and take half of 6 and square it

add these numbers to both sides of the equation

x^2-4x+4+y^2+6y+9=3+4+9

(x^2-4x+4)+(y^2+6y+9)=16

(x-2)^2+(y+3)^2=4^2

center is (2,-3)

radius is 4