Page 61 - TUTORIAL MATEMATIKA IPA KELAS XI-2

LINGKARAN

9. ∴ Pusat (-1, 1), r = 2
2
2
2
2
2
→ ( − (−1)) + ( − 1) = 2 → ( + 1) + ( − 1) = 4
2
2
→ + + 2 − 2 − 2 = 0

2
2
10. ∴ (2, 2) → 2 + 2 + . 2 + . 2 + = 0 → 2 + 2 + = −8 … ( )
2
2
∴ (2, 4) → 2 + 4 + . 2 + . 4 + = 0 → 2 + 4 + = −20 … ( )
2
2
∴ (5, −1) → 5 + (−1) + . 5 + . −1 + = 0 → 5 − + = −26 … ( )
Eliminasi (i) dan (ii) → = −6
Eliminasi (i) dan (iii) → −3 + 3 = 18 → −3 + 3. −6 = 18 → −3 − 18 = 18 → = −12
Dari (i) → 2. −12 + 2. −6 + = −8 → −24 − 12 + = −8 → = 28
2
2
Persamaan lingkaran → + − 12 − 6 + 28 = 0

2
2
2
2
11. Eliminasi + − 4 = 0 dan + − 4 = 0 diperoleh −4 + 4 = 0 → = 1
2
2
∴ = 1 → 1 + − 4 = 0
2
→ = 3 → = ±√3
→ (1, −√3) , (1, √3)
2
2
Titik pusat (4, 0) ; = √(4 − 1) + (0 − √3) = √12
2
2
2
2
2
Persamaan lingkaran → ( − 4) + ( − 0) = (√12) → + − 16 + 4 = 0

12. ∴ ( , ), (3, 0)
2
2
2
2
∴ = 2 → √( − 0) + ( − 0) = 2√( − 3) + ( − 0)
2
2
2
2
2
2
2
2
→ + = 4( − 6 + 9 + ) → + = 4 + 4 − 24 + 36
2
2
→ 3 + 3 − 24 + 36 = 0

0
2( 10 cos )+3( ) ( 20 cos ) 4 cos
13. ∴ ∶ = 2 ∶ 3 → ( ) = 2 +3 = 10 sin 5 = 20 sin +15 = ( )
2+3 5 5 4 sin + 3

→ = 4 cos → cos =
4
−3
→ = 4 sin + 3 → sin =
4 2 2
2
2

2
2
→ (cos ) + (sin ) = 1 → ( ) + ( −3 ) = + −6 +9 = 1
4 4 16
2
2
2
2
→ + − 6 + 9 = 16 → + − 6 − 7 = 0

14. Jari-jari, r = 2 – (-3) = 5 ; titik pusat (1, 2)
2
2
2
2
2
Persamaan lingkaran ( − 1) + ( − 2) = 5 → + − 2 − 4 − 20 = 0

15. Jari-jari = jarak titik pusat (4, 4)ke garis + − 2 → = | 4.1+4.1−2 | = 6 = 3√2
2
√1 +1 2 √2
2
2
2
Persamaan lingkaran → ( − 4) + ( − 4) = (3√2)
2
2
→ + − 8 − 8 + 14 = 0 4 (4, 4)
x + y = 2 r
2

2 4

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