Roboguru

Salah satu persamaan garis singgung pada lingkaran  di titik yang berabsis  adalah ...

Pertanyaan

Salah satu persamaan garis singgung pada lingkaran begin mathsize 14px style open parentheses straight x minus 2 close parentheses squared plus open parentheses straight y plus 1 close parentheses squared equals 13 end style di titik yang berabsis begin mathsize 14px style negative 1 end style adalah ...

  1. begin mathsize 14px style 3 straight x plus 2 straight y plus 9 equals 0 end style 

  2. Error converting from MathML to accessible text. 

  3. Error converting from MathML to accessible text.

  4. undefined 

Pembahasan Soal:

Substitusi begin mathsize 14px style straight x equals negative 1 end style ke persamaan undefined

begin mathsize 14px style space space space space left parenthesis straight x minus 2 right parenthesis squared plus left parenthesis straight y plus 1 right parenthesis squared equals 13 space left parenthesis negative 1 minus 2 right parenthesis squared plus left parenthesis straight y plus 1 right parenthesis squared equals 13 space space space space space space left parenthesis negative 3 right parenthesis squared plus left parenthesis straight y plus 1 right parenthesis squared equals 13 space space space space space space space space space space space space space space space space space space space space left parenthesis straight y plus 1 right parenthesis squared equals 13 minus 9 space space space space space space space space space space space space space space space space space space space space left parenthesis straight y plus 1 right parenthesis squared equals 4 space space space space space space space space space space space space space space space space space space space space space space space space y plus 1 equals plus-or-minus 2 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 y equals negative 1 plus-or-minus 2 comma space sehingga 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 straight y equals negative 1 minus 2 space space space space space space space space space straight y equals negative 1 plus 2 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 straight y equals negative 3 space space space space space space space space space space space space space space straight y equals 1 end style 

Koordinat titik singgungnya adalah begin mathsize 14px style open parentheses negative 1 comma negative 3 close parentheses space dan space open parentheses negative 1 comma 1 close parentheses end style.

Persamaan garis singgungnya 

begin mathsize 14px style open parentheses straight x subscript 1 minus 2 close parentheses open parentheses straight x minus 2 close parentheses plus open parentheses straight y subscript 1 plus 1 close parentheses open parentheses straight y plus 1 close parentheses equals 13 end style

Substitusikan koordinat titik singgungnya ke persamaan garis singgung di atas.

  • Untuk begin mathsize 14px style open parentheses negative 1 comma negative 3 close parentheses end style

Error converting from MathML to accessible text. 

  • Untuk begin mathsize 14px style open parentheses negative 1 comma 1 close parentheses end style

begin mathsize 14px style open parentheses negative 1 minus 2 close parentheses open parentheses straight x minus 2 close parentheses plus open parentheses 1 plus 1 close parentheses open parentheses straight y plus 1 close parentheses equals 13 space space space space space space space space space space space space space space space space minus 3 open parentheses straight x minus 2 close parentheses plus 2 open parentheses straight y plus 1 close parentheses equals 13 space space space space space space space space space space space space space space space space space space space space space minus 3 straight x plus 6 plus 2 straight y plus 2 equals 13 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 minus 3 straight x plus 2 straight y minus 5 equals 0 end style

Jadi, jawaban yang tepat adalah A.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 02 Februari 2021

Ruangguru

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved