what is the center of a circle whose equation is $x^{2}+y^{2}-12x - 2y+12 = 0$?\n(-12, -2)\n(-6, -1)\n(6…

what is the center of a circle whose equation is $x^{2}+y^{2}-12x - 2y+12 = 0$?\n(-12, -2)\n(-6, -1)\n(6, 1)\n(12, 2)

what is the center of a circle whose equation is $x^{2}+y^{2}-12x - 2y+12 = 0$?\n(-12, -2)\n(-6, -1)\n(6, 1)\n(12, 2)

Answer

Explanation:

Step1: Rewrite the equation in standard form

Complete the square for (x) and (y) terms. For (x) - terms: (x^{2}-12x=(x - 6)^{2}-36). For (y) - terms: (y^{2}-2y=(y - 1)^{2}-1). The original equation (x^{2}+y^{2}-12x - 2y+12 = 0) becomes ((x - 6)^{2}-36+(y - 1)^{2}-1 + 12=0).

Step2: Simplify the equation

((x - 6)^{2}+(y - 1)^{2}-36-1 + 12=0), which simplifies to ((x - 6)^{2}+(y - 1)^{2}=25). The standard - form of a circle equation is ((x - a)^{2}+(y - b)^{2}=r^{2}), where ((a,b)) is the center of the circle.

Answer:

C. ((6,1))