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Diketahui dua lingkaran dengan persamaan  dan , tentukan: d. Kedudukan kedua lingkaran tersebut.

Pertanyaan

Diketahui dua lingkaran dengan persamaan x squared plus y squared plus 5 x minus 3 y minus 14 equals 0 dan x squared plus y squared plus 4 x minus 2 y minus 12 equals 0, tentukan:

d. Kedudukan kedua lingkaran tersebut.

Pembahasan Soal:

Pada lingkaran dengan persamaan x squared plus y squared plus A x plus B y plus C equals 0 diketahui bahwa:

a equals negative 1 half A b equals negative 1 half B r squared equals a squared plus b squared minus C

atau

r equals square root of A squared over 4 plus B squared over 4 minus C end root

dimana left parenthesis a comma b right parenthesis adalah titik pusat lingkaran, dan r adalah panjang jari-jari lingkaran.

  • Pada persamaan lingkaran x squared plus y squared plus 5 x minus 3 y minus 14 equals 0.

A equals 5 B equals negative 3 C equals negative 14

 r equals square root of left parenthesis 5 right parenthesis squared over 4 plus left parenthesis negative 3 right parenthesis squared over 4 minus left parenthesis negative 14 right parenthesis end root space equals space square root of 25 over 4 plus 9 over 4 plus 14 end root space equals square root of 34 over 4 plus 14 end root space equals square root of 34 over 4 plus 56 over 4 end root space equals square root of 90 over 4 end root space equals fraction numerator square root of 90 over denominator square root of 4 end fraction space equals fraction numerator 3 square root of 10 over denominator 2 end fraction

Jadi, panjang jari-jari persamaan lingkaran tersebut adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank r end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 3 over 2 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 10 end cell end table.

  • Pada persamaan lingkaran x squared plus y squared plus 4 x minus 2 y minus 12 equals 0.

A equals 4 B equals negative 2 C equals negative 12

r equals square root of left parenthesis 4 right parenthesis squared over 4 plus left parenthesis negative 2 right parenthesis squared over 4 minus left parenthesis negative 12 right parenthesis end root space equals space square root of 16 over 4 plus 4 over 4 plus 12 end root space equals square root of 20 over 4 plus 12 end root space equals square root of 20 over 4 plus 48 over 4 end root space equals square root of 68 over 4 end root space equals square root of 17  

Jadi, panjang jari-jari persamaan lingkaran tersebut adalah r equals square root of 17.

Untuk menentukan kedudukan kedua lingkaran akan dilihat hubungan P subscript 1 P subscript 2 dan r subscript 1 plus r subscript 2.

table attributes columnalign right center left columnspacing 0px end attributes row cell P subscript 1 P subscript 2 end cell equals cell square root of open parentheses x subscript 2 minus x subscript 1 close parentheses squared plus open parentheses y subscript 2 minus y subscript 1 close parentheses squared end root end cell row blank equals cell square root of open parentheses negative 2 minus open parentheses negative 5 over 2 close parentheses close parentheses squared plus open parentheses 1 minus 3 over 2 close parentheses squared end root end cell row blank equals cell square root of open parentheses fraction numerator negative 4 plus 5 over denominator 2 end fraction close parentheses squared plus open parentheses fraction numerator 2 minus 3 over denominator 2 end fraction close parentheses squared end root end cell row blank equals cell square root of open parentheses 1 half close parentheses squared plus open parentheses negative 1 half close parentheses squared end root end cell row blank equals cell square root of 1 fourth plus 1 fourth end root end cell row blank equals cell square root of 2 over 4 end root end cell row blank equals cell 1 half square root of 2 end cell row blank equals cell 0 comma 5 open parentheses 1 comma 4 close parentheses end cell row blank equals cell 0 comma 7 end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell r subscript 1 plus r subscript 2 end cell equals cell 3 over 2 square root of 10 plus square root of 17 end cell row blank equals cell 1 comma 5 open parentheses 3 comma 16 close parentheses plus 4 comma 12 end cell row blank equals cell 4 comma 74 plus 4 comma 12 end cell row blank equals cell 8 comma 86 end cell end table 

Dengan demikian, karena table attributes columnalign right center left columnspacing 0px end attributes row cell P subscript 1 P subscript 2 end cell less than cell r subscript 1 plus r subscript 2 end cell end table maka kedua lingkaran berpotongan di dua titik.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 02 Juni 2021

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