answer:To determine the number of common tangents between two circles, we need to consider the position relationship between the circles. Firstly, we will calculate the distance between the centers ( O_1 ) and ( O_2 ): d = sqrt{[(2 - (-1)]^{2} + [(-3) - 1]^{2}} = sqrt{(2 + 1)^{2} + (-3 - 1)^{2}} = sqrt{3^2 + (-4)^2} = sqrt{9 + 16} = sqrt{25} = 5. Next, we find the radii of the two circles: For Circle O_1, the radius ( r_1 ) is the square root of 4, ( r_1 = 2 ). For Circle O_2, the radius ( r_2 ) is the square root of 9, ( r_2 = 3 ). The sum of the radii of the two circles is ( r_1 + r_2 = 2 + 3 = 5 ), which is equal to the distance between their centers d. Since the distance between the centers of two circles ( d ) is equal to the sum of their radii ( r_1 + r_2 ), the circles are externally tangent to each other. For two externally tangent circles, there are three common tangents: 1. One external tangent that touches both circles on the outside. 2. Two 'cross' tangents that each touch both circles, one on each circle. Therefore, there are (boxed{3}) common tangents between circles ( O_1 ) and ( O_2 ).

answer:(The full score for this question is 12 points) Solution: (1) Since c(acos B- frac {1}{2}b)=a^{2}-b^{2}. By the cosine rule, we can get: a^{2}+c^{2}-b^{2}-bc=2a^{2}-2b^{2}. It follows that: a^{2}=c^{2}+b^{2}-bc, Therefore, cos A= frac {b^{2}+c^{2}-a^{2}}{2bc}= frac {1}{2}, Since Ain(0,pi), Therefore, A= frac {pi}{3}…6 points (2) sin B+sin C=sin B+sin (A+B)=sin B+sin Acos B+cos Asin B = frac {3}{2}sin B+ frac { sqrt {3}}{2}cos B= sqrt {3}sin (B+ frac {pi}{6}), Since Bin(0, frac {2pi}{3}), Therefore, B+ frac {pi}{6}in( frac {pi}{6}, frac {5pi}{6}), sin (B+ frac {pi}{6})in( frac {1}{2},1], Therefore, the maximum value of sin B+sin C is boxed{sqrt {3}}…12 points

answer:Since f(x) = begin{cases} -x-4, & (x < 0) x^{2}-4, & (x > 0) end{cases}, we have begin{cases} x < 0 -x-4=0 end{cases} or begin{cases} x > 0 x^{2}-4=0 end{cases}. Solving these gives: x=-4, or x=2. The zeros of the function are: -4, 2; Therefore, the correct answer is: boxed{text{B}}. By using the piecewise function and solving f(x)=0, we can find the solution. This question tests the solution of the analytical expression of a piecewise function and the solution of the zeros of a function, and is considered a basic question.

answer:(I) The function f(x)=2cos (2x+ frac {π}{3})-2cos 2x+1 can be simplified as follows: begin{align} f(x) &= 2cos (2x+ frac {π}{3})-2cos 2x+1 &= - cos 2x - sqrt{3} sin 2x + 1 &= - 2 sin (2x + frac{π}{6}) + 1 end{align} The center of symmetry of the graph of the function f(x) satisfies the equation 2x + frac{π}{6} = kπ, where k in mathbb{Z}. Therefore, x = -frac{π}{12} + frac{k}{2}π. Thus, the center of symmetry is boxed{(-frac{π}{12} + frac{k}{2}π, 1)}. (II) Since f(A) = 0, we have: begin{align} -2 sin (2A + frac{π}{6}) + 1 &= 0 sin (2A + frac{π}{6}) &= frac{1}{2} 2A + frac{π}{6} &= frac{5π}{6} A &= frac{π}{3} end{align} Since triangle ABC is an acute-angled triangle, we know that B + C = frac{2π}{3}. Now, let's find the ratio frac{b}{c}: frac{b}{c} = frac{sin B}{sin C} = frac{frac{sqrt{3}}{2} cos C + frac{1}{2} sin C}{sin C} = frac{sqrt{3}}{2} frac{1}{tan C} + frac{1}{2} Since 0 < B < frac{π}{2} and 0 < C < frac{π}{2}, we have frac{π}{6} < C < frac{π}{2}, which implies that tan C > sqrt{3}. Consequently, frac{1}{tan C} < frac{1}{sqrt{3}}, and thus frac{1}{2} < frac{b}{c} < 2. Hence, the range of values for boxed{frac{b}{c}} is (frac{1}{2}, 2).