diberikan persamaan lingkaran x2+y2+ax_by+C=0 melalui titik''(1,2) (1,2) (2,1 dan (1,0)nilai dari (a+b+c) adalah

jawab

x²+y² + ax - by + c = 0

(1,2) → 1 + 4 + a - 2b + c = 0
a - 2b + c = - 5 ...(i)

(2,1) → 4 + 1 + 2a - b + c = 0
2a - b + c = - 5...(ii)

(1,0) → 1 + 0 + a - c = 0
a - c = -1 ...(iii)

(i) dan (ii)
a - 2b + c =  -5
2a - b + c = - 5...}kali 2

a - 2b + c = - 5
4a - 2b + 2c = -10...||(-)
-3a - c = 5...(iv)

(iv dan (iii)
- 3a - c  = 5
a - c = - 1...(-)
-4a = 6
a = - 6/4 = - 3/2
c = a + 1
c = -3/2 + 1 = - 1/2

a - 2b + c = -5
-3/2 - 2b  -1/2 = 5
-2 - 2b = 5
2b = - 7
b = - 7/2

a+ b+ c = -3/2  - 7/2 -1/2 = -11/2