diket
pusat (1, -2)
a,b (2 , 3)
ditanya persm lingkaran
jawab :
(x-a)² + (y-b)² = r²
(1 - 2)² + ( -2 - 3)² = r²
(-3)² + (-5)² = r²
9 + 25 = r²
r² = 36
r = 6
prsm lingkaran
(x-a)² + (y-b)² =r²
(x-2)² + (y-3)² = 36
(x² - 4x + 4 ) + (y² - 6y + 9) = 36
x² + y² - 4x -6y + 4 + 9 - 36 = 0
x² + y² - 4x - 6y - 23 = 0
cmiiw :)
Jawab:✓(∆x)²+(∆y)²=√( 1-2)²+(-2-3)²
=√(-1)²+(-5)²
=√1+25=√26
Pers O berpusat (a,b)
(x-a)²+(y-b)²=r²
(x-(-2)²+(y-3)²=√26²
(x+2)²+(y-3)²=26
x²+4x+4+y²-6y+9=26
x²+y²+4x-6y+9+4-26=0
x²+y²+4x-6y-13=0
Penjelasan dengan langkah-langkah:
" Life is not a problem to be solved but a reality to be experienced! "
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diket
pusat (1, -2)
a,b (2 , 3)
ditanya persm lingkaran
jawab :
(x-a)² + (y-b)² = r²
(1 - 2)² + ( -2 - 3)² = r²
(-3)² + (-5)² = r²
9 + 25 = r²
r² = 36
r = 6
prsm lingkaran
(x-a)² + (y-b)² =r²
(x-2)² + (y-3)² = 36
(x² - 4x + 4 ) + (y² - 6y + 9) = 36
x² + y² - 4x -6y + 4 + 9 - 36 = 0
x² + y² - 4x - 6y - 23 = 0
cmiiw :)
Jawab:✓(∆x)²+(∆y)²=√( 1-2)²+(-2-3)²
=√(-1)²+(-5)²
=√1+25=√26
Pers O berpusat (a,b)
(x-a)²+(y-b)²=r²
(x-(-2)²+(y-3)²=√26²
(x+2)²+(y-3)²=26
x²+4x+4+y²-6y+9=26
x²+y²+4x-6y+9+4-26=0
x²+y²+4x-6y-13=0
Penjelasan dengan langkah-langkah: