Circle

a)
C1:x2+y2-4x+6y+8 = 0
Centre of C1: (-(-4/2), -6/2) = (2, -3)

C2: x2+y2-10x-6y+14 = 0
Centre of C2: (-(-10/2), -(-6/2)) = (5, 3)

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b)
C1:x2+y2-4x+6y+8 = 0 ...... (1)
C2: x2+y2-10x-6y+14 = 0 ...... (2)

Consider the intersection between C1 and C2:
(1)-(2):
6x+12y-6 = 0
x+2y-1 = 0
x = -2y+1 ...... (3)

Put (3) into (1):
(-2y+1)2+y2-4(-2y+1)+6y+8 = 0
4y2-4y+1+y2+8y-4+6y+8 = 0
5y2+10y+5 = 0
y2 + 2y + 1 = 0

Determinant ∆
= b2 - 4ac
= (2)2 - 4(1)(1)
= 0

There are equal roots.
The two circles meet at one point.
Thus, the two circles touch each other.

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c)
From b):
C1:x2+y2-4x+6y+8 = 0 ...... (1)
C2: x2+y2-10x-6y+14 = 0 ...... (2)

x = -2y+1 ...... (3)

Put (3) into (1):
(-2y+1)2+y2-4(-2y+1)+6y+8 = 0
4y2-4y+1+y2+8y-4+6y+8 = 0
5y2+10y+5 = 0
y2 + 2y + 1 = 0
(y+1)2 = 0
y = -1

Put y = -1 into (3):
x = -2(-1)+1
x = 3

The point of the contact = (3. -1)

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d)
Slope of the line joining the two centres
= (-3-3)/(2-5)
= 2

Since the tangent ^ the line joining the two centres,
slope of the tangent = -(1/2)

Equation of the required tangent:
y-(-1) = -(1/2)(x-3)
x + 2y - 1 = 0
=