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Pertanyaan

Tentukan persamaan garis singgung yang melalui titik open parentheses 5 comma space 1 close parentheses pada lingkaran straight L identical to straight x squared plus straight y squared minus 4 straight x plus 6 straight y minus 12 equals 0.

W. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sriwijaya

Jawaban terverifikasi

Jawaban

 persamaan garis singgung yang melalui titik open parentheses 5 comma space 1 close parentheses pada lingkaran straight L identical to straight x squared plus straight y squared minus 4 straight x plus 6 straight y minus 12 equals 0 adalah 3 straight x plus 4 straight y minus 19 equals 0.

Pembahasan

Diketahui lingkaran straight L identical to straight x squared plus straight y squared minus 4 straight x plus 6 straight y minus 12 equals 0, maka:

straight A equals negative 4 comma space straight B equals 6 comma space straight C equals negative 12 

Sehingga persamaan garis singgung yang melalui titik open parentheses 5 comma space 1 close parentheses pada lingkaran straight L identical to straight x squared plus straight y squared minus 4 straight x plus 6 straight y minus 12 equals 0. adalah:

space space space space space space space space straight x subscript 1 times straight x plus straight y subscript 1 times straight y plus straight A times fraction numerator straight x subscript 1 plus straight x over denominator 2 end fraction plus straight B times fraction numerator straight y subscript 1 plus straight y over denominator 2 end fraction plus straight C equals 0 5 times straight x plus 1 times straight y plus open parentheses negative 4 close parentheses times fraction numerator 5 plus straight x over denominator 2 end fraction plus 6 times fraction numerator 1 plus straight y over denominator 2 end fraction plus open parentheses negative 12 close parentheses equals 0 space space space space space space space space space space space space space space space space space space space 5 straight x plus straight y minus 2 open parentheses 5 plus straight x close parentheses plus 3 open parentheses 1 plus straight y close parentheses minus 12 equals 0 space space space space space space space space space space space space space space space space space space space space space space 5 straight x plus straight y minus 10 minus 2 straight x plus 3 plus 3 straight y minus 12 equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 3 straight x plus 4 straight y minus 19 equals 0 

Jadi, persamaan garis singgung yang melalui titik open parentheses 5 comma space 1 close parentheses pada lingkaran straight L identical to straight x squared plus straight y squared minus 4 straight x plus 6 straight y minus 12 equals 0 adalah 3 straight x plus 4 straight y minus 19 equals 0.

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