Matematika

Pertanyaan

tentukan pusat dan jari-jari lingkaran :
~ x2+y2-8x+6y-11=0

1 Jawaban

  • x² + y² - 8x + 6y - 11 = 0

    p = [tex](- \frac{1}{2}A, -\frac{1}{2}B) = (- \frac{1}{2}(-8), -\frac{1}{2}.6) = (4, -3)[/tex]

    r = [tex] \sqrt{ \frac{1}{4}A^{2} + \frac{1}{4}B^{2} - C } \\ \sqrt{ \frac{1}{4}(-8)^{2} + \frac{1}{4}(6)^{2} + 11} \\ \sqrt{ \frac{1}{4}.64 + \frac{1}{4}.36 + 11 } \\ \sqrt{16 + 9 + 11} \\ \sqrt{36} = 6 [/tex]

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