Let's solve the problem step-by-step.
### Part (a): Finding the Two Possible Values of [tex]\( p \)[/tex]
Given:
1. Line equation: [tex]\( 2x + y - 5 = 0 \)[/tex]
2. Circle equation: [tex]\( (x - 3)^2 + (y - p)^2 = 5 \)[/tex]
We know that for a line [tex]\(Ax + By + C = 0\)[/tex] to be tangent to a circle with center [tex]\((x_0, y_0)\)[/tex] and radius [tex]\(r\)[/tex], the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle.
The perpendicular distance [tex]\(D\)[/tex] from a point [tex]\((x_0, y_0)\)[/tex] to the line [tex]\(Ax + By + C = 0\)[/tex] is given by:
[tex]\[
D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
\][/tex]
For the given circle, the center [tex]\((x_0, y_0)\)[/tex] is [tex]\((3, p)\)[/tex] and the radius [tex]\(r\)[/tex] is [tex]\(\sqrt{5}\)[/tex].
For the given line [tex]\(2x + y - 5 = 0\)[/tex]:
- [tex]\(A = 2\)[/tex]
- [tex]\(B = 1\)[/tex]
- [tex]\(C = -5\)[/tex]
We need this distance to be equal to the radius [tex]\(\sqrt{5}\)[/tex]:
[tex]\[
\frac{|2 \cdot 3 + 1 \cdot p - 5|}{\sqrt{2^2 + 1^2}} = \sqrt{5}
\][/tex]
Simplify the equation step-by-step:
1. Substitute [tex]\(x_0 = 3\)[/tex] and [tex]\(y_0 = p\)[/tex]:
[tex]\[
\frac{|6 + p - 5|}{\sqrt{4 + 1}} = \sqrt{5}
\][/tex]
2. Simplify inside the absolute value:
[tex]\[
\frac{|1 + p|}{\sqrt{5}} = \sqrt{5}
\][/tex]
3. Multiply both sides by [tex]\(\sqrt{5}\)[/tex]:
[tex]\[
|1 + p| = 5
\][/tex]
4. Solve the absolute value equation:
- Case 1: [tex]\(1 + p = 5\)[/tex]
[tex]\[
p = 5 - 1 = 4
\][/tex]
- Case 2: [tex]\(1 + p = -5\)[/tex]
[tex]\[
p = -5 - 1 = -6
\][/tex]
Thus, the two possible values for [tex]\(p\)[/tex] are [tex]\(4\)[/tex] and [tex]\(-6\)[/tex].
### Part (b): Coordinates of the Centre of the Circle
For each value of [tex]\(p\)[/tex]:
1. When [tex]\(p = 4\)[/tex], the center of the circle is [tex]\((3, 4)\)[/tex].
2. When [tex]\(p = -6\)[/tex], the center of the circle is [tex]\((3, -6)\)[/tex].
Hence, the centers of the circle in each case are:
- [tex]\((3, 4)\)[/tex]
- [tex]\((3, -6)\)[/tex]
