Page 65 - TUTORIAL MATEMATIKA IPA SEMESTER 2

LINGKARAN

2
2
2
30. = 2 → + 2 − 2 − 2.2 + 1 = 0 → − 2 + 1 = 0
2
= (−2) − 4.1.1 = 0 → menyinggung lingkaran

31. Titik pusat (1, −1) = (− , − )
2 2

→ 1 = − → = −2 ; −1 = − → = 2
2 2
2
2
2
= → + − 2 + 2 + = 0 → 2 + = 0
Agar garis dan lingkaran bersinggungan maka D = 0.
2
→ 0 − 4.2. = 0 → −8 = 0 → = 0
p + q + r = -2 + 2 + 0 = 0

2
2
2
2
32. = 2 → 2 = √( ) − → 4 = − → = − 4 … ( )
2
2
2
2
2
− = 0 → = → + − 2 + = 0 → 2 − 2 + − 4 = 0
Agar garis dan lingkaran bersinggungan maka D = 0.
2
2
2
2
2
→ (−2 ) − 4.2. ( − 4) = 0 → 4 − 8 + 32 = 0 → −4 = −32
2
→ = 8 → = √8 = 2√2

2 2
2
2
33. + ( 1 (2 + 5)) − 4 − = 0 → + (2 +5) − 4 − = 0
√5 5
2
2
2
5 +4 +20 +25−20 −5 9 +25−5 2
→ = 0 → = 0 → 9 + 25 − 5 = 0
5 5
Agar garis dan lingkaran bersinggungan maka D = 0.
2
→ 0 = 0 − 4.9. (25 − 5 ) → 0 = 0 − 900 + 180 → 180 = 900 → = 5

34. + = → = −
2
2
2
2
2
2
2
( − ) + = 1 − → − 2 + + − 1 + = 0 → 2 − 2 + + − 1 = 0
Agar garis dan lingkaran berpotongan maka D > 0.
2
2
2
2
→ (−2 ) − 4.2. ( + − 1) > 0 → 4 − 8 − 8 + 8 > 0
2
2
2
→ −4 − 8 + 8 > 0 → + 2 − 2 < 0 ; 1,2 = −2±√2 −4.1.(−2) = −1 ± √3
2.1
→ = −1 + √3 , = −1 − √3
2
1

++++++ -------------------- +++++++
-1−√3 -1+√3 HP = { | − 1 − √3 < < −1 + √3, ∈ }

35. Titik (1, 1) terletak pada lingkaran.
4 −2
2
2
Titik pusat (− , − ) = (−2, 1) ; = √(−2) + (1) − (−4) = 3
2 2
2
PGS : (1 − (−2))( − (−2)) + (1 − 1)( − 1) = 3 → 3 + 6 = 9 → = 1

2
2
2
36. = −1 → + (−1) + 4 − 2. (−1) − 15 = 0 → + 4 − 12 = 0 → ( + 6)( − 2) = 0
Terdapat dua titik yang dilalui garis singgung (-6, -1) dan (2, -1).
4 −2
2
2
Titik pusat (− , − ) = (−2, 1) ; = √(−2) + 1 − (−15) = √20
2 2
PGS 1 melalui titik (−6, −1)
→ (−6 − (−2))( − (−2)) + (−1 − 1)( − 1) = 20 → 4 + 2 = −26 → 2 + = −13
PGS 2 melalui titik (2, −1)
→ (2 − (−2))( − (−2)) + (−1 − 1)( − 1) = 20 → 4 + 2 = 10 → 2 + = 5
LEARNING IS FUN 64