Question

Nilai n agar persamaan lingkaran
x2+y2+2x+10y+n=0

mempunyai jari – jari 8 adalah …

Answer 1

X²+y²+2x+10y+n = 0
x²+2x+1-1+y²+10y+25-25+n=0
(x+1)²-1+(y+5)²-25+n = 0
(x+1)²+(y+5)²-26+n=0
(x+1)+(y+5) = (26-n)--->r²

26-n = r²
26-n = 8²
26-n = 64
n = 26-64 = -34

Answer 2

N = - 38

x^2 + y^2 + 2× + 10y - 38 = 0
×^2 + 2× + 1 + y^2 + 10y + 25 = 38 + 1 + 25
(× + 1)^2 + (y + 5)^2 = 64

maka, pusatnya (-1, -5) dan r = 8
karena r^2 = 64
r = akar 64
r= 8