Cosθ=cos2∠OPQ=3/5。
2.( 1) Prove that the distance from point C (1, 2) to line L is d = 3m+ 1 √ (5m? +6m+2)r = 5
Alternatively, according to the straight line L, you can pass the fixed point A (3, 1) because it is in a circle.
(2) When the shortest length of the line segment of line 1 is cut by circle C, it is given by Pythagorean theorem dmax, and d = 3m+ 1 √ (5m?
+6m+2)=√(9m？ +6m+ 1)/(5m？ +6m+2)= √[ 9/5-(24/5m+ 13/5)/5m？ +6m+2]
Let 24/5m+ 13/5=t, then m=(5t- 13)/24, then d=√[9/5-576t/( 125t? +70t+125)] = √ [9/5-576/﹙125t+125/t+70 ﹚] dmax is obvious when t￠0, that is, D.
3.( 1) From the meaning of the question, f(x)=X? +2x+b has only one intersection A(0, b) on the y axis, so there must be two intersections x 1, x2 on the x axis, which is equivalent to the equation x? +2x+b=0 has two real roots, δ = 4-4b-0, so b < 1.
(2) Let the equation of a circle be x? +y？ +Dx+Ey+F=0, let x=0, y? +Dy+F=0, what about b? +Eb+F=0, let y=0, x? +Dx+F=0, x 1, x2 is the equation x? +2x+b = the root of 0, and x? +dx+f = the root of 0, so D=2, F=b, so E=-b- 1, and the equation of the circle is x? +y？ +2x-(b+ 1)y+b=0
(3) adopt the parameter set, b( 1-y)+x? +y？ +2x-y=0, let 1-y=0, that is, y= 1, then x? +2x=0, x=0 or -2, so the fixed point is (0, 1) or (-2, 1).