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)

Answer:

the center the circle is (6,1)  

Step-by-step explanation: