Tentukan persamaan garis singgung pada lingkaran x2+y2=25 yang melalui titik T(7, 0) di luar lingkaran!

Pertanyaan

Tentukan persamaan garis singgung pada lingkaran begin mathsize 14px style x squared plus y squared equals 25 end style yang melalui titik begin mathsize 14px style straight T open parentheses 7 comma space 0 close parentheses end style di luar lingkaran!

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Jawaban terverifikasi

Jawaban

 persamaan garis singgung pada lingkaran begin mathsize 14px style x squared plus y squared equals 25 end style yang melalui titik begin mathsize 14px style straight T open parentheses 7 comma space 0 close parentheses end style di luar lingkaran adalah begin mathsize 14px style y equals 5 over 12 square root of 6 x minus 35 over 12 square root of 6 end style dan begin mathsize 14px style y equals negative 5 over 12 square root of 6 x plus 35 over 12 square root of 6 end style.

Pembahasan

Persamaan garis singgung yang melalui titik begin mathsize 14px style straight T open parentheses 7 comma space 0 close parentheses end style adalah:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell y minus y subscript 1 end cell equals cell m open parentheses x minus x subscript 1 close parentheses end cell row cell y minus 0 end cell equals cell m open parentheses x minus 7 close parentheses end cell row y equals cell m x minus 7 m end cell end table end style 

Dengan substitusi persamaan garis di atas ke dalam persamaan lingkaran diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus y squared end cell equals 25 row cell x squared plus open parentheses m x minus 7 m close parentheses squared end cell equals 25 row cell x squared plus m squared x squared minus 14 m squared x plus 49 m squared end cell equals 25 row cell open parentheses 1 plus m squared close parentheses x squared minus 14 m squared x plus 49 m squared minus 25 end cell equals 0 end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Diperoleh colon space a end cell equals cell 1 plus m squared end cell row b equals cell negative 14 m squared end cell row c equals cell 49 m squared minus 25 end cell end table end style 

Karena garis tersebut menyinggung lingkaran, maka diskriminan bernilai nol sehingga diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row D equals 0 row cell b squared minus 4 a c end cell equals 0 row cell open parentheses negative 14 m squared close parentheses squared minus 4 open parentheses 1 plus m squared close parentheses open parentheses 49 m squared minus 25 close parentheses end cell equals 0 row cell 196 m to the power of 4 minus 4 open parentheses 49 m squared minus 25 plus 49 m to the power of 4 minus 25 m squared close parentheses end cell equals 0 row cell 196 m to the power of 4 minus 4 open parentheses 49 m to the power of 4 plus 24 m squared minus 25 close parentheses end cell equals 0 row cell 196 m to the power of 4 minus 196 m to the power of 4 minus 96 m squared plus 100 end cell equals 0 row cell negative 96 m squared plus 100 end cell equals 0 end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 96 m squared end cell equals 100 row cell m squared end cell equals cell 100 over 96 end cell row blank equals cell 25 over 24 end cell row m equals cell plus-or-minus square root of 25 over 24 end root end cell row blank equals cell plus-or-minus fraction numerator 5 over denominator 2 square root of 6 end fraction cross times fraction numerator 2 square root of 6 over denominator 2 square root of 6 end fraction end cell row blank equals cell plus-or-minus 5 over 12 square root of 6 end cell end table end style 

Untuk begin mathsize 14px style m equals 5 over 12 square root of 6 end style diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell m x minus 7 m end cell row y equals cell open parentheses 5 over 12 square root of 6 close parentheses x minus 7 open parentheses 5 over 12 square root of 6 close parentheses end cell row y equals cell 5 over 12 square root of 6 x minus 35 over 12 square root of 6 end cell end table end style 

Untuk begin mathsize 14px style m equals negative 5 over 12 square root of 6 end style diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell m x minus 7 m end cell row y equals cell open parentheses negative 5 over 12 square root of 6 close parentheses x minus 7 open parentheses negative 5 over 12 square root of 6 close parentheses end cell row y equals cell negative 5 over 12 square root of 6 x plus 35 over 12 square root of 6 end cell end table end style 

Jadi, persamaan garis singgung pada lingkaran begin mathsize 14px style x squared plus y squared equals 25 end style yang melalui titik begin mathsize 14px style straight T open parentheses 7 comma space 0 close parentheses end style di luar lingkaran adalah begin mathsize 14px style y equals 5 over 12 square root of 6 x minus 35 over 12 square root of 6 end style dan begin mathsize 14px style y equals negative 5 over 12 square root of 6 x plus 35 over 12 square root of 6 end style.

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zalfa zahirah

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Leni Ardiwanti zil

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Tentukan persamaan garis singgung dari titik (0, 6) pada lingkaran x2+y2=25.

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