The center of the circle q is on the positive half axis of X, the line l1:3x + 4y-8 = 0 is tangent to the circle Q, and the line L2 with the slope k passing through the point P (0,2) intersects the circle Q at two points a and B The radius of the circle is 2 The equation for finding circle Q Finding the value range of real number k Is there a constant k such that the vector OA + ob is collinear with PQ? If there is a constant K, there is no reason

The center of the circle q is on the positive half axis of X, the line l1:3x + 4y-8 = 0 is tangent to the circle Q, and the line L2 with the slope k passing through the point P (0,2) intersects the circle Q at two points a and B The radius of the circle is 2 The equation for finding circle Q Finding the value range of real number k Is there a constant k such that the vector OA + ob is collinear with PQ? If there is a constant K, there is no reason

(1) Let the center of the circle be (a, 0) (a > 0) on the positive half axis of X, and let the radius be equal to 2 and tangent to the line L1. Using the distance from the center of the circle to the line equal to the radius, the Q equation of the circle is obtained as follows: (X-6) & sup2; + Y & sup2; = 4;
(2) Let the line be y = KX + 2, then because the line intersects the circle, the distance from the center of the circle to the line is less than the radius, so it is - 3 / 4