the straight line 2x-y+8=0 and the circle 2x^2+2y^2-3x+2y+k=0 do not have any intersection points,where k is a constant.find the range of the values of k

Step 1: Rewrite the equation of the line in slope-intercept form (y = mx + b). The equation 2x - y + 8 = 0 can be rewritten as y = 2x + 8.

Step 2: Substitute y = 2x + 8 into the equation of the circle. This gives us 2x^2 + 2(2x + 8)^2 - 3x + 2(2x + 8) + k = 0.

Step 3: Simplify the equation. This gives us 10x^2 - 4x + 32 + k = 0.

Step 4: The discriminant of a quadratic equation ax^2 + bx + c = 0 is given by b^2 - 4ac. For the quadratic to have no real roots (i.e., for the line and circle to have no intersection points), the discriminant must be less than 0.

Step 5: Calculate the discriminant of the equation 10x^2 - 4x + 32 + k = 0. This gives us (-4)^2 - 410(32 + k) = 16 - 40*(32 + k).

Step 6: Set the discriminant less than 0 and solve for k. This gives us 16 - 40*(32 + k) < 0, which simplifies to k > -30.

Therefore, the range of values for k is k > -30.