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Lana71 [14]
2 years ago
10

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

Mathematics
2 answers:
Marianna [84]2 years ago
7 0

Answer:

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

Step-by-step 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 = \sqrt{25} = 5

Olin [163]2 years ago
3 0

Answer:

center of the circle = (2, -3)

radius of the circle = 5

Step-by-step explanation:

<u>Equation of a circle</u>

(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
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