answer:1. **Application of the Power of a Point Theorem**: - Since AB is the diameter and AD a chord, the segment DF where F is the extended point hitting the tangent, and AF intersecting AB at A. - Let DF = p and AD = q. Since AB = 2r, apply the power of a point theorem: [ BF cdot BA = DF^2 implies BF cdot 2r = p^2. ] - We can solve for BF: [ BF = frac{p^2}{2r}. ] 2. **Use the condition AG = GF**: - As AG = GF, and GF = BF - FD, it follows AG = BF - p. - Substitute for BF: [ AG = frac{p^2}{2r} - p. ] 3. **Connecting u and v to AG and AB**: - By definition, u is the distance from G to the tangent at A, similar to AG. - v is the perpendicular distance from G to the diameter AB, which remains constant at r (radius of the circle). 4. **Formulate the equation**: - Given u = AG = frac{p^2}{2r} - p and v = r, establish the relationship. - Replace p = r - u into AG: [ u = frac{(r-u)^2}{2r} - (r-u). ] - Expand and simplify to find v^2 in terms of u: [ v^2 = r^2 = frac{u^3}{2r-u}. ] Thus, the correct relation between u and v is: [ v^2 = frac{u^3{2r-u}} ] The final answer is boxed{A}

answer:**Analysis** This question examines the positional relationship between a line and a parabola, which is a basic question. Pay attention to the special case where the line is parallel to the axis of the parabola. **Solution** Solution: When the line is parallel to the axis of the parabola, that is, when the equation of the line is in the form of y=a, the line and the parabola have only one common point but do not touch each other. Therefore, the given statement is incorrect. Thus, the correct choice is boxed{text{B}}.

answer:1. **Identify the equations of the lines and parabola**: - From (2x-2y+4)(6x+2y-8)=0: - 2x-2y+4=0 Rightarrow y = x + 2 - 6x+2y-8=0 Rightarrow y = -3x + 4 - From (2x + 2y - 4)=0 text{ and } y = -x^2+2: - 2x + 2y - 4 = 0 Rightarrow y = -x + 2 - Parabola: y = -x^2 + 2 2. **Check for intersections**: - **Intersection of y = x + 2 with y = -x + 2**: [ x + 2 = -x + 2 Rightarrow 2x = 0 Rightarrow x = 0 Rightarrow y = 2 ] - **Intersection of y = x + 2 with y = -x^2 + 2**: [ x + 2 = -x^2 + 2 Rightarrow x^2 + x = 0 Rightarrow x(x+1) = 0 Rightarrow x = 0, -1 ] Thus, (y = 2) when (x = 0) and (y = 1) when (x = -1). - **Intersection of y = -3x + 4 with y = -x + 2**: [ -3x + 4 = -x + 2 Rightarrow -2x = -2 Rightarrow x = 1 Rightarrow y = -1 ] - **Intersection of y = -3x + 4 with y = -x^2 + 2** is solved by substituting (y = -3x + 4) into the parabola equation and solving for (x). [ -3x + 4 = -x^2 + 2 Rightarrow x^2 - 3x + 2 = 0 Rightarrow (x-2)(x-1) = 0 Rightarrow x = 2, 1 ] Thus, (y = 2) when (x = 2) and (-1) when (x = 1). But this repeats an earlier solution point. 3. **Conclusion with boxed answer**: - The distinct points of intersection are (0, 2), (-1, 1), (1, -1), text{and } (2, 2). Therefore, the number of points common to the graphs of the modified equations is 4. The final answer is boxed{B) 4}

answer:Solution: (I) Since overrightarrow{m}perp overrightarrow{n}, we have overrightarrow{m}cdot overrightarrow{n}=sin ^{2}B-sin ^{2}C+sin ^{2}A-sin Asin B=0, By the Law of Sines, we get b^{2}-c^{2}+a^{2}-ab=0, which simplifies to a^{2}+b^{2}-c^{2}=ab, Using the Law of Cosines, we find cos C= frac {a^{2}+b^{2}-c^{2}}{2ab}= frac {ab}{2ab}= frac {1}{2}. Therefore, since 0 < C < pi, we have C= frac {pi}{3}. (II) Hence, sin C= frac { sqrt {3}}{2}= frac { sqrt {75}}{10} > frac { sqrt {64}}{10}= frac {4}{5}=sin A, So, by the Law of Sines, we know c > a, then frac {pi}{3}=C > A, thus cos A= frac {3}{5}, Therefore, cos B=-cos (A+C)=sin Asin C-cos Acos C= frac {4 sqrt {3}-3}{10}. Thus, the answers are: (I) The magnitude of angle C is boxed{frac {pi}{3}}. (II) The value of cos B is boxed{frac {4 sqrt {3}-3}{10}}.