Equation of circle

(a) By the fact that AP = BP:
(h - 4)2 + (k - 5)2 = (h - 7)2 + (k - 4)2
-8h + 16 - 10k + 25 = -14h + 49 - 8k + 16
-2k = -6h + 24
k = 3h - 12
(b) Equation of C:
(x - h)2 + (y - k)2 = (h - 7)2 + (k - 4)2
x2 + y2 - 2hx - 2ky + h2 + k2 = h2 + k2 - 14h - 8k + 65
x2 + y2 - 2hx - 2ky = - 14h - 8k + 65
x2 + y2 - 2hx - 2ky + 14h + 8k - 65 = 0
x2 + y2 - 2hx - 2(3h - 12)y + 14h + 8(3h - 12) - 65 = 0
x2 + y2 - 2hx - (6h - 24)y + (38h - 161) = 0
(c) (i) The tangent equation is:
7x + 4y - h(x + 7) - (3h - 12)(y + 4) + (38h - 161) = 0
7x + 4y - hx - 7h - 3hy - 12h + 12y + 48 + 38h - 161 = 0
(7 - h)x + (16 - 3h)y + (19h - 113) = 0
(ii) Slope of tangent at B = (h - 7)/(16 - 3h)
Slope of PA = (k - 5)/(h - 4) = (3h - 17)/(h - 4)
Therefore:
(3h - 17)/(h - 4) = (h - 7)/(16 - 3h)
(3h - 17)(16 - 3h) = (h - 7)(h - 4)
-9h2 + 99h - 272 = h2 - 11h + 28
10h2 - 110h + 300 = 0
h2 - 11h + 30 = 0
(h - 6)(h - 5) = 0
h = 5 or 6

(h-4)+(k-5)=(h-7)+(k-4) (By distance formula)
h-8h+16+k-10k+25=h-14h+49+k-8k+16
-8h+16-10k+25=-14h+49-8k+16
2k=6h-24
k=3h-12
∴Centre of the circle=(h,3h-12)
∴Raduis of circle=sqrt{(h-4)+[(3h-12)-5]}
=sqrt(h-8h+16+9h-102h+289)
=sqrt(10h-110h+305)
Radius =sqrt[(D/2)+(E/2)-F]
∴h+(3h-12)-F=10h-110h+305
10h-72h+144-F=10h-110h+305
∴F=38h-161
∴Equation of C is x +y -2hx -(6h-24)y +(38h-161) = 0---------(1)
Let the equation of tangent of C at B be y-4=m(x-7)
(By calculation, you can get the answer)
Slope of PA=(k-5)/(h-4)
=(3h-17)/(h-4)
∴(3h-17)/(h-4)=(h-7)/(16-3h)
-(3h-17)(3h-16)=(h-7)(h-4)
-9h+99h-272=h-11h+28
10h-110h+300=0
h-11h+30=0
(h-5)(h-6)=0
h=5 or 6