Question

Find the center of a circle with the equation: x2+y2−2y−8x−19=0

Answer 1

its (-4,1) r6

Explanation:

Answer 2

The equation of a circle in Center-Radius form is given by :


`(x-a)^(2) + (y-b)^(2) = R^(2)`, where the center of the circle is the point (a, b) and the radius is R.

To find the center and radius of x2+y2−2y−8x−19=0 we write this equation again in Center-Radius form by completing the square:


`x^(2) + y^(2) -2y-8x-19=0`


`x^(2)-8x + y^(2) -2y-19=0`


`(x^(2)-2*4x +16)-16 + (y^(2) -2*1y+1)-1-19=0`


`(x-4)^(2)+(y-1)^(2)=16+1+19`


`(x-4)^(2)+(y-1)^(2)=36`


`(x-4)^(2)+(y-1)^(2)= 6^(2)`

Thus, the center of the radius is (4, 1) and radius is 6


Answer: (4,1)