Question

Determine the central and radius of the circle of equation given below
x2+y2-4x+6y-12=0

Answer 1

centre = (2, - 3 ) and radius = 5

Explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r is the radius

given

x² + y² - 4x + 6y - 12 = 0 ( add 12 to both sides )

x² + y² - 4x + 6y = 12 ( collect x/ y terms )

x² + 4x + y² + 6y = 12

using the method of completing the square

add ( half the coefficient of the x/ y terms )² to both sides

x² + 2(-2)x + 4 + y² + 2(3)y + 9 = 12 + 4 + 9

(x - 2)² + (y + 3)² = 25 ← in standard form

with (h, k ) = ( 2, - 3 ) and r =
`√(25)` = 5

Answer 2

center of the circle = (2, -3)

radius of the circle = 5

Explanation:

Equation of a circle

`(x-a)^2+(y-b)^2=r^2`

(where (a, b) is the center and r is the radius)

Given equation:

`x^2+y^2-4x+6y-12=0`

Collect like terms:

`\implies x^2-4x+y^2+6y-12=0`

Add 12 to both sides:

`\implies x^2-4x+y^2+6y=12`

Complete the square for both variables.

Add 4 to both sides for x. Add 9 to both sides for y.

`\implies x^2-4x+4+y^2+6y+9=12+4+9`

`\implies (x^2-4x+4)+(y^2+6y+9)=12+4+9`

Factor the two variables:

`\implies (x-2)^2+(y+3)^2=25`

Therefore:

• center of the circle = (2, -3)
• radius of the circle = √25 = 5