Q:

What is the center of a circle whose equation is x^2+y^2-12x-2y+12=0

Accepted Solution

A:
Answer:$$(6,\, 1)$$.Step-by-step explanation:Let $$(a,\, b)$$ denote the center of this circle (where $$a$$ and $$b$$ are to be found.) If the radius of this circle is $$r$$, the equation of this circle would be:$$(x - a)^{2} + (y - b)^{2} = r^{2}$$.Expand this equation using binomial theorem:$$x^{2} - 2\, a\, x + a^{2} + y^{2} - 2\, b\, y + b^{2} = r^{2}$$.Both this equation and the given equation $$x^{2} + y^{2} - 12\, x - 2\, y + 12 = 0$$  describe the same circle. Therefore, corresponding coefficients of the two equations should match one another:The coefficient of the $$x$$ term should match in the two equations. Therefore: $$(-2\, a) = (-12)$$.The coefficient of the $$y$$ term should also match in the two equations. Therefore: $$(-2\, b) = (-2)$$.Solve these two equation for $$a$$ and $$b$$: $$a = 6$$ and $$b = 1$$.Substitute $$a = 6$$ and $$b = 1$$ into $$(a,\, b)$$ to find the center of this circle: $$(6,\, 1)$$.