Very Urgent...please help me to solve this CE a.maths...thz

(a)
Since the equation of the line PQ is C2-C1
(x^2 + y^2 + 10x + 4y + F) - ( x^2 + y^2 + 4x - 2y + 1)
6x+6y+(F-1)
But PQ is x+y+3=0 (given)
So 6x+6y+(F-1)=x+y+3
F-1=18
F=19
(b)
We know that if there is a circle x^2+y^2+Dx+Ey+F=0
and M(x1,y1) is a external point, then the distance of the tangent between M and the circle is
√(x1^2+y1^2+Dx1+Ey1+F)
So the length of the tangent from M to C1 D1 is

√(x1^2+y1^2+4x1-2y1+1)
The length of the tangent from M to C2 D2 is
√(x1^2+y1^2+10x1+4y1+19)
Consider D2^2-D1^2
(x1^2+y1^2+10x1+4y1+19)-(x1^2+y1^2+4x1-2y1+1)
=6x1+6y1+18
=6(x1+y1+3)
=0 ( Since (x1,y1) on the line PQ)
We prove that D1^2=D2^2
That is D1=D2
So the length of the tangent from M to C1 is equal to the length of tangent from M to C2.