Tentukan pusat dan jaring jaring dari persamaan lingkaran berikut1.) x²+y²-12x-4y+36=0
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Penjelasan dengan langkah-langkah:
x² + y² - 12x - 4y + 36 = 0
a = -12, b = -4, c = 36
pusat = [-½a, -½b]
pusat = [-½(-12), -½(-4)]
pusat = [6, 2]
[tex]\sf radius = \sqrt{ \frac{1}{2} {a}^{4} + \frac{1}{4} {b}^{2} - c } [/tex]
[tex]\sf radius = \sqrt{ \frac{1}{4} ( - 12) {}^{2} + \frac{1}{4} ( - 4) {}^{2} - 36} [/tex]
[tex]\sf radius = \sqrt{ \frac{1}{4}(144) + \frac{1}{4} (16) - 36 } [/tex]
[tex]\sf radius = \sqrt{36 + 4 - 36} [/tex]
[tex]\sf radius = \sqrt{40 - 36} [/tex]
[tex]\sf radius = \sqrt{4} [/tex]
[tex]\sf radius =2[/tex]
