Roboguru

Titik Pusat dan jari-jari lingkaran yang mempunyai persamaan adalah...

Pertanyaan

Titik Pusat dan jari-jari lingkaran yang mempunyai persamaan x squared plus y squared equals 81 adalah...

  1. O left parenthesis 0 comma 0 right parenthesis spacedan r equals 3

  2. O left parenthesis 0 comma 0 right parenthesis spacedan r equals 6

  3. O left parenthesis 0 comma 0 right parenthesis spacedan r equals 8

  4. O left parenthesis 0 comma 0 right parenthesis spacedan r equals 9

  5. O left parenthesis 0 comma 0 right parenthesis spacedan r equals 81

Pembahasan Soal:

x squared plus y squared equals 81

Terdapat aturan,

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus y squared end cell equals cell r squared end cell end table

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus y squared end cell equals cell r squared end cell row r equals cell square root of 81 end cell row r equals 9 end table

Oleh karena itu, jawaban  yang benar adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

G. Albiah

Mahasiswa/Alumni Universitas Galuh Ciamis

Terakhir diupdate 05 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Persamaan garis singgung lingkaran  yang sejajar dengan garis  adalah ...

Pembahasan Soal:

Persamaan lingkaran begin mathsize 14px style open parentheses x minus a close parentheses squared plus open parentheses y minus b close parentheses squared equals r squared end style mempunyai titik pusat begin mathsize 14px style open parentheses a comma space b close parentheses end style dengan jar-jari begin mathsize 14px style r end style.

Jika diketahui begin mathsize 14px style open parentheses x minus 3 close parentheses squared plus open parentheses y plus 5 close parentheses squared equals 80 end style maka diperoleh titik pusat lingkaran:

begin mathsize 14px style open parentheses a comma space b close parentheses equals open parentheses 3 comma space minus 5 close parentheses end style 

dan jari-jari:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell r squared end cell equals 80 row r equals cell square root of 80 end cell row blank equals cell square root of 16 times square root of 5 end cell row blank equals cell 4 square root of 5 end cell end table end style 

Persamaan garis singgung sejajar dengan garis begin mathsize 14px style y minus 2 x plus 5 equals 0 end style, karena sejajar maka mempunyai gradien yang sama sehingga:

begin mathsize 14px style y equals m x plus c y equals 2 x minus 5 end style 

diperoleh gradien begin mathsize 14px style m equals 2 end style.

Persamaan garis singung lingkaran berpusat di titik undefined dengan jari-jari undefined:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses y minus b close parentheses end cell equals cell m open parentheses x minus a close parentheses plus-or-minus r square root of m squared plus 1 end root end cell row cell open parentheses y plus 5 close parentheses end cell equals cell 2 open parentheses x minus 3 close parentheses plus-or-minus 4 square root of 5 times square root of 2 squared plus 1 end root end cell row cell open parentheses y plus 5 close parentheses end cell equals cell 2 open parentheses x minus 3 close parentheses plus-or-minus 4 square root of 5 times square root of 5 end cell row cell y plus 5 end cell equals cell 2 x minus 6 plus-or-minus 20 end cell row y equals cell 2 x minus 11 plus-or-minus 20 end cell end table end style 

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Tentukan pusat dan jari-jari pada lingkaran dengan persamaan berikut! d.

Pembahasan Soal:

Jika persamaan lingkaran berbentuk begin mathsize 14px style x squared plus y squared plus A x plus B y plus C equals 0 end style, maka titik pusatnya begin mathsize 14px style open parentheses negative 1 half A comma space minus 1 half B close parentheses end style dan jari-jarinya begin mathsize 14px style square root of 1 fourth A squared plus 1 fourth B squared minus C end root end style.

Sehingga dari persamaan lingkaran begin mathsize 14px style x squared plus y squared minus 4 x plus 4 y plus 4 equals 0 end style dapat diperoleh titik pusat dan jari-jarinya sebagai berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell Titik space pusat end cell equals cell open parentheses negative 1 half A comma space minus 1 half B close parentheses end cell row blank equals cell open parentheses negative 1 half open parentheses negative 4 close parentheses comma space minus 1 half open parentheses 4 close parentheses close parentheses end cell row blank equals cell open parentheses 2 comma space minus 2 close parentheses end cell end table end style  

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row r equals cell square root of 1 fourth A squared plus 1 fourth B squared minus C end root end cell row blank equals cell square root of 1 fourth open parentheses negative 4 close parentheses squared plus 1 fourth open parentheses 4 close parentheses squared minus open parentheses 4 close parentheses end root end cell row blank equals cell square root of 1 fourth open parentheses 16 close parentheses plus 1 fourth open parentheses 16 close parentheses minus 4 end root end cell row blank equals cell square root of 4 plus 4 minus 4 end root end cell row blank equals cell square root of 4 end cell row blank equals 2 end table end style 

Jadi, diperoleh titik pusat begin mathsize 14px style left parenthesis 2 comma space minus 2 right parenthesis end style dan jari-jarinya begin mathsize 14px style 2 end style.

0

Roboguru

Tentukan titik pusat dan jari-jari lingkaran .

Pembahasan Soal:

Bentuk umum persamaan lingkaran:

begin mathsize 14px style x squared plus y squared plus A x plus B y plus C equals 0 end style

titik pusat: begin mathsize 14px style P left parenthesis a comma b right parenthesis equals P open parentheses negative 1 half A comma negative 1 half B close parentheses end style

jari-jari: begin mathsize 14px style r equals square root of a squared plus b squared minus C end root end style 

Persamaan lingkaran

 begin mathsize 14px style x squared plus y squared plus 4 x minus 7 y minus 40 equals 0 p u s a t equals open parentheses negative 1 half cross times 4 comma negative 1 half cross times open parentheses negative 7 close parentheses close parentheses space space space space space space space space space equals open parentheses negative 2 comma 7 over 2 close parentheses j a r i minus j a r i equals square root of open parentheses negative 2 close parentheses squared plus open parentheses 7 over 2 close parentheses squared minus open parentheses negative 40 close parentheses end root space space space space space space space space space space space space space space equals square root of 4 plus 49 over 4 plus 40 end root space space space space space space space space space space space space space space equals square root of 44 plus 49 over 4 end root space space space space space space space space space space space space space space equals space square root of fraction numerator 176 plus 49 over denominator 4 end fraction end root space space space space space space space space space space space space space space equals square root of 225 over 4 end root space space space space space space space space space space space space space space equals 15 over 2 end style 

Jadi, titik pusat lingkaran tersebut adalah begin mathsize 14px style open parentheses negative 2 comma 7 over 2 close parentheses end style dan jari-jarinya adalah begin mathsize 14px style 15 over 2 end style.

0

Roboguru

Persamaan lingkaran yang sepusat dengan lingkaran lain yang memiliki persamaan  dan berjari-jari  satuan adalah ....

Pembahasan Soal:

Diketahui persamaan begin mathsize 14px style x squared plus y squared minus 2 x plus 6 y plus 12 equals 0 end style, maka:

begin mathsize 14px style straight P open parentheses fraction numerator straight A over denominator negative 2 end fraction comma space fraction numerator straight B over denominator negative 2 end fraction close parentheses equals straight P open parentheses fraction numerator negative 2 over denominator negative 2 end fraction comma space fraction numerator 6 over denominator negative 2 end fraction close parentheses equals straight P open parentheses 1 comma space minus 3 close parentheses end style 

sehingga persamaan lingkaran yang sepusat dan berjari-jari begin mathsize 14px style 8 end style satuan dapat di tentukan dengan cara:

- Pilihan A dengan persamaan size 14px x to the power of size 14px 2 size 14px plus size 14px y to the power of size 14px 2 size 14px minus size 14px 2 size 14px x size 14px plus size 14px 6 size 14px y size 14px plus size 14px 54 size 14px equals size 14px 0 

begin mathsize 14px style straight P open parentheses fraction numerator straight A over denominator negative 2 end fraction comma space fraction numerator straight B over denominator negative 2 end fraction close parentheses equals straight P open parentheses fraction numerator negative 2 over denominator negative 2 end fraction comma space fraction numerator 6 over denominator negative 2 end fraction close parentheses equals straight P open parentheses 1 comma space minus 3 close parentheses straight r equals square root of open parentheses fraction numerator straight A over denominator negative 2 end fraction close parentheses squared plus open parentheses fraction numerator straight A over denominator negative 2 end fraction close parentheses squared minus straight C end root equals square root of 1 squared plus open parentheses negative 3 close parentheses squared minus 54 end root equals square root of negative 44 end root end style 

- Pilihan B dengan persamaan begin mathsize 14px style x squared plus y squared minus 2 x plus 6 y minus 54 equals 0 end style 

begin mathsize 14px style straight P open parentheses fraction numerator straight A over denominator negative 2 end fraction comma space fraction numerator straight B over denominator negative 2 end fraction close parentheses equals straight P open parentheses fraction numerator negative 2 over denominator negative 2 end fraction comma space fraction numerator 6 over denominator negative 2 end fraction close parentheses equals straight P open parentheses 1 comma space minus 3 close parentheses straight r equals square root of open parentheses fraction numerator straight A over denominator negative 2 end fraction close parentheses squared plus open parentheses fraction numerator straight A over denominator negative 2 end fraction close parentheses squared minus straight C end root equals square root of 1 squared plus open parentheses negative 3 close parentheses squared plus 54 end root equals square root of 64 equals 8 end style 

- Pilihan C dengan persamaan begin mathsize 14px style x squared plus y squared plus 2 x minus 6 y minus 54 equals 0 end style 

begin mathsize 14px style straight P open parentheses fraction numerator straight A over denominator negative 2 end fraction comma space fraction numerator straight B over denominator negative 2 end fraction close parentheses equals straight P open parentheses fraction numerator 2 over denominator negative 2 end fraction comma space fraction numerator negative 6 over denominator negative 2 end fraction close parentheses equals straight P open parentheses negative 1 comma space 3 close parentheses straight r equals square root of open parentheses fraction numerator straight A over denominator negative 2 end fraction close parentheses squared plus open parentheses fraction numerator straight A over denominator negative 2 end fraction close parentheses squared minus straight C end root equals square root of open parentheses negative 1 close parentheses squared plus open parentheses 3 close parentheses squared plus 54 end root equals square root of 64 equals 8 end style 

- Pilihan D dengan persamaan begin mathsize 14px style x squared plus y squared minus 6 x plus 2 y plus 54 equals 0 end style 

begin mathsize 14px style straight P open parentheses fraction numerator straight A over denominator negative 2 end fraction comma space fraction numerator straight B over denominator negative 2 end fraction close parentheses equals straight P open parentheses fraction numerator negative 6 over denominator negative 2 end fraction comma space fraction numerator 2 over denominator negative 2 end fraction close parentheses equals straight P open parentheses 3 comma space minus 1 close parentheses straight r equals square root of open parentheses fraction numerator straight A over denominator negative 2 end fraction close parentheses squared plus open parentheses fraction numerator straight A over denominator negative 2 end fraction close parentheses squared minus straight C end root equals square root of 3 squared plus open parentheses negative 1 close parentheses squared minus 54 end root equals square root of negative 44 end root end style 

- Pilihan E dengan persamaan begin mathsize 14px style x squared plus y squared minus 6 x plus 2 y minus 54 equals 0 end style 

begin mathsize 14px style straight P open parentheses fraction numerator straight A over denominator negative 2 end fraction comma space fraction numerator straight B over denominator negative 2 end fraction close parentheses equals straight P open parentheses fraction numerator negative 6 over denominator negative 2 end fraction comma space fraction numerator 2 over denominator negative 2 end fraction close parentheses equals straight P open parentheses 3 comma space minus 1 close parentheses straight r equals square root of open parentheses fraction numerator straight A over denominator negative 2 end fraction close parentheses squared plus open parentheses fraction numerator straight A over denominator negative 2 end fraction close parentheses squared minus straight C end root equals square root of 3 squared plus open parentheses negative 1 close parentheses squared plus 54 end root equals square root of 64 equals 8 end style 

Jadi, jawaban yang benar adalah B.

0

Roboguru

Tentukan titik pusat dan jari-jari dari persamaan lingkaran berikut: a.  .

Pembahasan Soal:

Menentukan titik pusat dan jari-jari lingkaran.

Bentuk Umum persamaan lingkaran.

begin mathsize 14px style straight x squared plus straight y squared plus Ax plus By plus straight C equals 0 end style 

maka:

begin mathsize 14px style Pusat equals left parenthesis negative 1 half straight A comma negative 1 half straight B right parenthesis straight r equals square root of left parenthesis negative 1 half straight A right parenthesis squared plus left parenthesis negative 1 half straight B right parenthesis squared minus straight C end root end style 

Diketahui pada soal: persamaan lingkaran begin mathsize 14px style straight x squared plus straight y squared minus 4 straight x plus 8 straight y minus 5 equals 0 end style

begin mathsize 14px style straight A equals negative 4 straight B equals 8 straight C equals negative 5 end style 

Maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row Pusat equals cell left parenthesis negative 1 half left parenthesis negative 4 right parenthesis comma negative 1 half times 8 right parenthesis end cell row blank equals cell left parenthesis 2 comma space minus 4 right parenthesis end cell row straight r equals cell square root of 2 squared plus left parenthesis negative 4 right parenthesis squared minus left parenthesis negative 5 right parenthesis end root end cell row blank equals cell square root of 4 plus 16 plus 5 end root end cell row blank equals cell square root of 25 end cell row blank equals 5 end table end style 

Jadi, Titik pusat dan jari-jari dari persamaan lingkaran begin mathsize 14px style straight x squared plus straight y squared minus 4 straight x plus 8 straight y minus 5 equals 0 end style adalah begin mathsize 14px style left parenthesis 2 comma space minus 4 right parenthesis end style dan 5.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

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