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pusat (0, 0) dan r = √32 = 4√2
2x + y = 4
y = -2x + 4 → m = -2
sejajar → m1 = m = -2
y = m1x ± r√(1 + m²)
y = -2x ± 4√2√(1 + (-2)²)
y = -2x + 4√10, atau y = -2x - 4√10
2. tegak lurus → m1 = -1/m
m1 = -1/-2 = ½
y = ½x ± 4√2√(1 + (½)²)
y = ½x ± 4√(10/4)
y = ½x ± 2√10, atau dikali2
PGS
2y = x + 4√10 atau 2y = x - 4√10