Question

The equation for the circle is:

x2+y2−12x+6y−19=0 .

What is the center of the circle?
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Answer 1

$x^{2} - 12x + y^{2} + 6y = 19$
$x^{2} - 12x + 36 + y^{2} + 6y + 9 = 19 + 36 + 9$
$(x-6)^{2} + (y+3)^{2} = 64$

Center: (6,-3)

Answer 2

The centre of the circle with the equation x²+ y²−12x + 6y − 19 = 0 are 6 and -3

How to find the centre of a circle?

The equation of the circle is represented as follows;

x²+ y²−12x + 6y − 19 = 0

Therefore, the equation of a circle is represented as follows:

(x - h)² + (y - k)² = r²

where

h and k are the centre.

Hence,

x²+ y²−12x + 6y − 19 = 0

x² - 12x  + y² + 6y - 19 = 0

(x- 6)² + y² + 6y = 19

Therefore,

(x-6)² + (y+3)² = 64

Therefore, the centre are 6 and -3

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