Question

find the centre of the circle 4x^2+4y^2+16x- 24y-52=0​

Answer 1

Answer:

(- 2, 3 )

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 the radius

Given

4x² + 4y² + 16x - 24y - 52 = 0 ( divide through by 4 )

x² + y² + 4x - 6y - 13 = 0 ← add 13 to both sides and rearrange terms on left side

x² + 4x + y² - 6y = 13

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 = 13 + 4 + 9

(x + 2)² + (y - 3)² = 26 ← in standard form

with centre (- 2, 3 )

Answer 2

Answer:

Centre at $(-2,3)$.

Step-by-step explanation:

Use the general formula for a circle to answer this question. The general formula is :

$(x-a)^2+(y-b)^2=r^2$, where $(a,b)$is the centre of the circle and $r$ is the radius of the circle.

We have to get the equation we've been given into the general form. We can do this by completing the square as follows:

$4x^2+16x+4y^2-24y-52=0$ -Group like terms together.

$x^2+4x+y^2-6y-13=0$ - Divide by common factor of 4.

$(x+2)^2+(y-3)^2=26$ - Complete the square and move constant to the other sides.

Now that it's in the general form we can find the centre. Centre at $(-2,3)$.