Respuesta :
The equation of a circle in Center-Radius form is given by :
[tex] (x-a)^{2} + (y-b)^{2} = R^{2} [/tex], 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:
[tex] x^{2} + y^{2} -2y-8x-19=0[/tex]
[tex] x^{2}-8x + y^{2} -2y-19=0[/tex]
[tex](x^{2}-2*4x +16)-16 + (y^{2} -2*1y+1)-1-19=0[/tex]
[tex] (x-4)^{2}+(y-1)^{2}=16+1+19[/tex]
[tex] (x-4)^{2}+(y-1)^{2}=36[/tex]
[tex] (x-4)^{2}+(y-1)^{2}= 6^{2} [/tex]
Thus, the center of the radius is (4, 1) and radius is 6
Answer: (4,1)
[tex] (x-a)^{2} + (y-b)^{2} = R^{2} [/tex], 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:
[tex] x^{2} + y^{2} -2y-8x-19=0[/tex]
[tex] x^{2}-8x + y^{2} -2y-19=0[/tex]
[tex](x^{2}-2*4x +16)-16 + (y^{2} -2*1y+1)-1-19=0[/tex]
[tex] (x-4)^{2}+(y-1)^{2}=16+1+19[/tex]
[tex] (x-4)^{2}+(y-1)^{2}=36[/tex]
[tex] (x-4)^{2}+(y-1)^{2}= 6^{2} [/tex]
Thus, the center of the radius is (4, 1) and radius is 6
Answer: (4,1)
