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Persamaan lingkaran yang melalui titik  adalah ...

Pertanyaan

Persamaan lingkaran yang melalui titik begin mathsize 14px style left parenthesis 3 comma space 5 right parenthesis comma space left parenthesis negative 4 comma space minus 2 right parenthesis comma space dan space left parenthesis 3 comma space minus 1 right parenthesis end style adalah ...

Pembahasan Soal:

Melalui titik begin mathsize 14px style left parenthesis 3 comma space 5 right parenthesis end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 3 squared plus 5 squared plus A times 3 plus B times 5 plus C end cell equals 0 row cell 9 plus 25 plus 3 A plus 5 B plus C end cell equals 0 row cell 3 A plus 5 B plus C plus 34 end cell equals 0 row cell 3 A plus 34 end cell equals cell negative open parentheses 5 B plus C close parentheses end cell row cell 3 A end cell equals cell negative 5 B minus C minus 34 end cell row A equals cell fraction numerator negative 5 B minus C minus 34 over denominator 3 end fraction end cell end table end style ...(1)

Melalui titik begin mathsize 14px style left parenthesis negative 4 comma space minus 2 right parenthesis end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared plus A left parenthesis negative 4 right parenthesis plus B left parenthesis negative 2 right parenthesis plus C end cell equals 0 row cell 16 plus 4 minus 4 A minus 2 B plus C end cell equals 0 row cell negative 4 A minus 2 B plus C plus 20 end cell equals 0 end table end style...(2)

Melalui begin mathsize 14px style left parenthesis 3 comma space minus 1 right parenthesis end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis 3 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared plus A left parenthesis 3 right parenthesis plus B left parenthesis negative 1 right parenthesis plus C end cell equals 0 row cell 9 plus 1 plus 3 A minus B plus C end cell equals 0 row cell 3 A minus B plus C plus 10 end cell equals 0 end table end style...(3)

Dengan metode substitusi, didapat: 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Substitusi thin space A end cell equals cell fraction numerator negative 5 B minus C minus 34 over denominator 3 end fraction end cell row cell negative 4 times fraction numerator negative 5 B minus C minus 34 over denominator 3 end fraction minus 2 B plus C plus 20 end cell equals cell 0... left parenthesis 2 right parenthesis end cell row cell negative fraction numerator 4 open parentheses negative 5 B minus C minus 34 close parentheses over denominator 3 end fraction minus 2 B plus C plus 20 end cell equals 0 row cell fraction numerator 14 B plus 7 C plus 136 over denominator 3 end fraction plus 20 end cell equals cell 0... left parenthesis 4 right parenthesis end cell row blank blank blank row cell 3 times fraction numerator negative 5 B minus C minus 34 over denominator 3 end fraction minus B plus C plus 10 end cell equals cell 0 space... left parenthesis 3 right parenthesis end cell row cell open parentheses negative 5 B minus C minus 34 close parentheses minus B plus C plus 10 end cell equals 0 row cell negative 5 B minus C minus 34 minus B plus C plus 10 end cell equals 0 row cell negative 6 B minus 24 end cell equals 0 row B equals cell negative 4 end cell row blank blank blank row cell Substitusi thin space B end cell equals cell negative 4 end cell row cell fraction numerator 14 open parentheses negative 4 close parentheses plus 7 C plus 136 over denominator 3 end fraction plus 20 end cell equals cell 0 space... left parenthesis 4 right parenthesis end cell row cell fraction numerator negative 14 times thin space 4 plus 7 C plus 136 over denominator 3 end fraction plus 20 end cell equals 0 row cell fraction numerator 7 C plus 80 over denominator 3 end fraction plus 20 end cell equals 0 row cell fraction numerator 7 C plus 80 over denominator 3 end fraction end cell equals cell negative 20 end cell row cell 7 C plus 80 end cell equals cell negative 60 end cell row cell 7 C end cell equals cell negative 140 end cell row C equals cell negative 20 end cell row blank blank blank row cell Substitusi thin space C end cell equals cell negative 20 comma thin space B equals negative 4 end cell row A equals cell fraction numerator negative 5 open parentheses negative 4 close parentheses minus open parentheses negative 20 close parentheses minus 34 over denominator 3 end fraction... left parenthesis 1 right parenthesis end cell row A equals cell fraction numerator 5 times thin space 4 plus 20 minus 34 over denominator 3 end fraction end cell row A equals cell 6 over 3 end cell row A equals 2 end table end style 

Didapat: begin mathsize 14px style A equals 2 comma thin space C equals negative 20 comma thin space B equals negative 4 end style.

Jadi, persamaan lingkarannya adalah begin mathsize 14px style x squared plus y squared plus 2 x minus 4 y minus 20 equals 0 end style.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Yoga

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 31 Maret 2021

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Pertanyaan yang serupa

Persamaan lingkaran berpusat di titik dan melalui titik adalah

Pembahasan Soal:

Diketahui titik pusat (2, 3) dan melalui titik (5, -1), maka a = 2, b = 3, c = 5, d = -1.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x minus a close parentheses squared plus open parentheses y minus b close parentheses squared end cell equals cell r squared end cell row cell open parentheses x minus a close parentheses squared plus open parentheses y minus b close parentheses squared end cell equals cell open parentheses c minus a close parentheses squared plus open parentheses d minus b close parentheses squared end cell row cell open parentheses x minus 2 close parentheses squared plus open parentheses y minus 3 close parentheses squared end cell equals cell open parentheses 5 minus 2 close parentheses squared plus open parentheses negative 1 minus 3 close parentheses squared end cell row cell x squared minus 4 x plus 4 plus y squared minus 6 y plus 9 end cell equals cell 9 plus 16 end cell row cell x squared plus y squared minus 4 x minus 6 y minus 12 end cell equals 0 end table end style

Jadi, persamaan lingkarannya adalah begin mathsize 14px style x squared plus y squared minus 4 x minus 6 y minus 12 equals 0 end style.

0

Roboguru

Lingkaran  melalui titik (1,7). Titik pusat lingkaran itu adalah

Pembahasan Soal:

Persamaan umum lingkaran

Berpusat di pangkal koordinat

begin mathsize 14px style x squared plus y squared equals r squared end style 

Berpusat di titik (a, b)

begin mathsize 14px style text (x-a) end text squared plus left parenthesis straight y minus straight b right parenthesis squared equals straight r to the power of 2 end exponent end style 

Diketahui :

Persamaan lingkaran: begin mathsize 14px style x squared plus y squared plus 4 x plus b y minus 12 equals 0 end style

Melalui titik : begin mathsize 14px style left parenthesis 1 comma 7 right parenthesis end style 

Ditanya :

Pusat lingkaran adalah

Jawab :

Langkah pertama kita subsitusikan titik (1, 7) kedalam persamaan lingkaran agar dapat nilai b

begin mathsize 14px style x squared plus y squared plus 4 x plus b y minus 12 equals 0 left parenthesis 1 right parenthesis squared plus left parenthesis 7 right parenthesis squared plus 4 left parenthesis 1 right parenthesis plus b left parenthesis 7 right parenthesis minus 12 equals 0 1 plus 49 plus 4 plus 7 b minus 12 equals 0 7 b plus 42 equals 0 7 b equals negative 42 b equals negative 42 over 7 b equals negative 6 end style 

Sehingga persamaan lingkarannya adalah begin mathsize 14px style x squared plus y squared plus 4 x minus 6 y minus 12 equals 0 end style 

Langkah selanjutnya kita cari pusat lingkaran

Persamaan Umum lingkaran begin mathsize 14px style x squared space plus space y squared space plus space A x plus B y plus C equals 0 end style, maka pusat lingkarannya adalah begin mathsize 14px style open parentheses negative 1 half A comma space minus 1 half B close parentheses end style.

Persamaan lingkaran: begin mathsize 14px style x squared plus y squared plus 4 x minus 6 y minus 12 equals 0 end style, maka pusat lingkarannya adalah begin mathsize 14px style equals open parentheses negative 1 half cross times 4 comma negative 1 half open parentheses negative 6 close parentheses close parentheses equals open parentheses negative 2 comma 3 close parentheses end style 

Jadi, jawaban yang paling benar adalah B.

0

Roboguru

Persamaan lingkaran yang berpusat di titk (2,4) dan melalui titik (10,-2) adalah ....

Pembahasan Soal:

Diketahui:

Pusat (2,4), melalui (10,-2)

Ditanya:

Persamaan lingkarannya adalah?

Jawab:

Rumus yang digunakan adalah

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row straight L identical to cell open parentheses straight x minus straight a close parentheses squared plus open parentheses straight y minus straight b close parentheses squared equals straight r squared end cell row blank identical to cell open parentheses straight x minus 2 close parentheses squared plus open parentheses straight y minus 4 close parentheses squared equals straight r squared end cell row blank blank blank end table end style 

Melalui titik (10,-2), kita substitusikan ke persamaan lingkaran

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 10 minus 2 close parentheses squared plus open parentheses negative 2 minus 4 close parentheses squared end cell equals cell straight r squared end cell row cell open parentheses 8 close parentheses squared plus open parentheses negative 6 close parentheses squared end cell equals cell straight r squared end cell row cell 64 plus 36 end cell equals cell straight r squared end cell row 100 equals cell straight r squared end cell row 10 equals straight r row blank blank blank row blank blank blank end table end style 

Jadi persamaan lingkarannya adalah 

begin mathsize 14px style open parentheses x minus 2 close parentheses squared plus open parentheses y minus 4 close parentheses squared equals 10 squared x squared minus 4 x plus 4 plus y squared minus 8 y plus 16 equals 100 x squared plus y squared minus 4 x minus 8 y minus 80 equals 0 end style 

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Persamaan Lingkaran dengan Titik Pusat  dan berjari-jari .

Pembahasan Soal:

Persamaan Lingkaran dengan Titik Pusat undefined dan berjari-jari undefined

Persamaan Lingkaran dengan Pusat (a,b) dan berjari-jari r

begin mathsize 14px style straight L identical to open parentheses x minus a close parentheses squared plus open parentheses y minus b close parentheses squared equals r squared straight L identical to open parentheses straight x minus 3 close parentheses squared plus open parentheses straight y minus 4 close parentheses squared equals 9 squared straight L identical to straight x squared minus 6 straight x plus 9 plus straight y squared minus 8 straight y plus 16 equals 81 straight L identical to straight x squared plus straight y squared minus 6 straight x minus 8 straight y minus 56 equals 0 end style 

0

Roboguru

Persamaan lingkaran yang berpusat di titik (3, - 4) dan berdiameter adalah....

Pembahasan Soal:

P e r s a m a a n space l i n g k a r a n colon space  left parenthesis x space minus space a right parenthesis ² space plus space left parenthesis y space minus space b right parenthesis ² space equals space r ² space  p u s a t space left parenthesis a comma b right parenthesis comma space j a r i minus j a r i space colon space r space space  p u s a t space left parenthesis 4 comma negative 3 right parenthesis space  d i a m e t e r space equals space 4 square root of 17  space j a r i minus j a r i space equals space r space equals space 2 square root of 17 space  left parenthesis x space minus space a right parenthesis ² space plus space left parenthesis y space minus space b right parenthesis ² space equals space r ²  left parenthesis x space minus space 4 right parenthesis ² space plus space left parenthesis y space plus space 3 right parenthesis ² space equals space left parenthesis 2 square root of 17 right parenthesis ²  space left parenthesis x space minus space 4 right parenthesis ² space plus space left parenthesis y space plus space 3 right parenthesis ² space equals space 68 space space    x ² space minus space 8 x space plus space 16 space plus space y ² space plus space 6 y space plus space 9 space equals space 68 space  x ² space plus space y ² space minus space 8 x space plus space 6 y space plus space 25 space minus space 68 space equals space 0 space  x ² space plus space y ² space minus space 8 x space plus space 6 y space minus space 43 space equals space 0 space space

0

Roboguru

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