Tentukan kuasa titik p
(1,3) pada lingkaran x2 + y2 – 2x – 4y -20. Tentukan pula
letak titik p terhadap lingkaran tersebut…..
Jawaban : X21
+ y21 + AX1 + By1 + C = 0
(1) + (3)2 – 2.1- 4 (3) – 20
= 0
1 + 9
– 2- 12 – 20 = 0
10
– 2- 12 – 20 = 0
-24
= 0
Tentuka persaman garis kuasa lingkaran
dimensi L1 = x2 + y2 = 25, L2 = (x2 + y2
– 2x – 4y- 4)
Jawaban :
L1 = x2 + y2
= 25
L2 = (x2 + y2 – 2x –
4y- 4)
L1 - L2 = 0
x2 + y2 - 25 - (x2
+ y2 – 2x – 4y- 4) = 0
x2 + y2 - 25 - x2
+ y2 + 2x + 4y + 4 = 0
2x + 4y
- 21= 0
L1 - L2 = 0
L1 = x2 + y2 + x +
y – 14 = 0
L2 = x2 + y2 = 13
L3 =
x2 + y2 + 3x – 2y – 26 = 0
L1 - L2 = 0
x2 + y2 + x + y –
14 - x2 + y2 -13 = 0
x + y – 14 – 13 = 0
x + y – 1
= 0
x
+ y – 1 = L3
x
+ y – 1 = x2 + y2 + 3x – 2y – 26 = 0
= x2 + y2 + 3x –
2y – 26 – x – y + 1
= x2 + y2 + 2x –
3y – 25
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