Final answer:
To find the values of p, use the distance from the center of the circle to the tangent line equation, equate it to the radius, and solve for p. The two possible values of p for the tangent condition are p = √(a) - 1 and p = -√(a) - 1.
Step-by-step explanation:
To find the two possible values of p where the line 2x+y-5=0 is tangent to the circle with equation (x-3)²+(y-p)²=5a, we'll need to use the condition that the distance from the center of the circle to the line equals the radius of the circle.
First, let's write the equation of the line in the slope-intercept form, which will give us y=-2x+5. The center of the circle is (3, p) and the radius of the circle is √(5a).
The distance d from a point (x1,y1) to a line Ax+By+C=0 is given by the formula d = |Ax1+By1+C| /√(A²+B²). For our case, the distance from the center of the circle (3, p) to the line y=-2x+5 is d = |2*3+p-5| /√(2²+1²), which simplifies to d = |1+p| /√5.
Since d must be equal to the radius √(5a) of the circle for the line to be tangent, we have the equation |1+p| /√5 = √(5a), or after squaring both sides (1+p)² = 5a. As a appears on both sides, we can cancel 5 out and are left with (1+p)² = a.
To solve for p, we take the square root of a to get two possible solutions, p = √(a) - 1 and p = -√(a) - 1. These are the two possible values for p where the line is tangent to the circle.