Respuesta :
The center of a circle is (6, 1)
Solution:
Given equation of circle is:
[tex]x^{2} +y^{2} -12x-2y+12=0[/tex]
Move the constant to other side
[tex]x^{2} +y^{2} -12x-2y = -12[/tex]
Group the equation
[tex](x^{2} -12x)+(y^{2}-2y)=-12[/tex]
Add 36 and 1 from both sides
[tex](x^{2} -12x +36)+(y^{2}-2y + 1)=-12 + 36 + 1\\\\(x^{2} -12x + 36)+(y^{2}-2y + 1)= 25[/tex]
Which is simplified as:
[tex](x - 6)^2 + (y-1)^2 = 25[/tex] --------- eqn 1
The equation of circle is given as:
[tex](x - h)^2 + (y - k)^2 = r^2[/tex]
Where, r is radius and (h, k) is center
Compare eqn 1 with above general equation
(h, k) = (6, 1)
Thus the center of a circle is (6, 1)