Iklan

Iklan

Pertanyaan

Tentukan persamaan lingkaran dengan pusat O : b. melalui titik ( 3 , 2 ) , ( − 2 , 7 ) , dan ( − 4 , − 3 ) .

Tentukan persamaan lingkaran dengan pusat :

b. melalui titik  .

Iklan

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

persamaan lingkaran yang melalui titik adalah

persamaan lingkaran yang melalui titik left parenthesis 3 comma space 2 right parenthesis comma space left parenthesis negative 2 comma space 7 right parenthesis comma space dan space left parenthesis negative 4 comma space minus 3 right parenthesis adalah Error converting from MathML to accessible text. 

Iklan

Pembahasan

Ingat kembali bentuk umum persamaan lingkaran yaitu Substitusikan ketiga titik tersebut untuk mencari nilai Melalui Melalui Melalui Gunakan eliminasi-substitusi untuk mencari nilai eliminasi eliminasi eliminasi Substitusikan nilai ke persamaan Substitusikan nilai ke bentuk umum persamaan lingkaran Jadi, persamaan lingkaran yang melalui titik adalah

Ingat kembali bentuk umum persamaan lingkaran yaitu x squared plus y squared plus a x plus b y plus c equals 0 

Substitusikan ketiga titik tersebut untuk mencari nilai a comma space b comma space dan space c 

Melalui left parenthesis 3 comma space 2 right parenthesis 

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus y squared plus a x plus b y plus c end cell equals 0 row cell left parenthesis 3 right parenthesis squared plus left parenthesis 2 right parenthesis squared plus a left parenthesis 3 right parenthesis plus b left parenthesis 2 right parenthesis plus c end cell equals 0 row cell 9 plus 4 plus 3 a plus 2 b plus c end cell equals 0 row cell 13 plus 3 a plus 2 b plus c end cell equals 0 row cell 3 a plus 2 b plus c end cell equals cell negative 13 space space space space space left parenthesis i right parenthesis end cell end table 

Melalui left parenthesis negative 2 comma space 7 right parenthesis 

 table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus y squared plus a x plus b y plus c end cell equals 0 row cell left parenthesis negative 2 right parenthesis squared plus left parenthesis 7 right parenthesis squared plus a left parenthesis negative 2 right parenthesis plus b left parenthesis 7 right parenthesis plus c end cell equals 0 row cell 4 plus 49 minus 2 a plus 7 b plus c end cell equals 0 row cell 53 minus 2 a plus 7 b plus c end cell equals 0 row cell negative 2 a plus 7 b plus c end cell equals cell negative 53 end cell row cell 2 a minus 7 b minus c end cell equals cell 53 space space space space space space space space space space left parenthesis i i right parenthesis end cell end table  

Melalui left parenthesis negative 4 comma space minus 3 right parenthesis 

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus y squared plus a x plus b y plus c end cell equals 0 row cell left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 3 right parenthesis squared plus a left parenthesis negative 4 right parenthesis plus b left parenthesis negative 3 right parenthesis plus c end cell equals 0 row cell 16 plus 9 minus 4 a minus 3 b plus c end cell equals 0 row cell 25 minus 4 a minus 3 b plus c end cell equals 0 row cell negative 4 a minus 3 b plus c end cell equals cell negative 25 end cell row cell 4 a plus 3 b minus c end cell equals cell 25 space space space space space space space space space space left parenthesis i i i right parenthesis end cell end table 

Gunakan eliminasi-substitusi untuk mencari nilai a comma space b comma space dan space c 

eliminasi left parenthesis i right parenthesis space dan space left parenthesis i i right parenthesis 

table row cell 3 a plus 2 b plus c equals negative 13 end cell row cell 2 a minus 7 b minus c equals 53 end cell row cell 5 a minus 5 b equals 40 space end cell row cell 5 left parenthesis a minus b right parenthesis equals 40 end cell row cell a minus b equals 8 space space left parenthesis i v right parenthesis end cell end table space plus    

eliminasi left parenthesis i right parenthesis space dan space left parenthesis i i i right parenthesis  

table row cell 3 a plus 2 b plus c equals negative 13 end cell row cell 4 a plus 3 b minus c equals 25 end cell row cell 7 a plus 5 b equals 12 space space space left parenthesis v right parenthesis end cell end table space plus 

eliminasi left parenthesis i v right parenthesis space dan space left parenthesis v right parenthesis 

 table row cell a minus b equals 8 end cell cell vertical line cross times 5 vertical line end cell cell 5 a minus 5 b equals 40 end cell row cell 7 a plus 5 b equals 12 end cell cell vertical line cross times 1 vertical line end cell cell 7 a plus 5 b equals 12 end cell row blank blank cell 12 a equals 52 end cell row blank blank cell a equals 52 over 12 end cell row blank blank cell a equals 13 over 3 end cell end table space plus  

table row cell a minus b equals 8 end cell cell vertical line cross times 7 vertical line end cell cell 7 a minus 7 b equals 56 end cell row cell 7 a plus 5 b equals 12 end cell cell vertical line cross times 1 vertical line end cell cell 7 a plus 5 b equals 12 end cell row blank blank cell negative 12 b equals 44 end cell row blank blank cell b equals negative 44 over 12 end cell row blank blank cell b equals negative 11 over 3 end cell end table space minus 

Substitusikan nilai a space dan space b ke persamaan left parenthesis i right parenthesis 

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 a plus 2 b plus c end cell equals cell negative 13 end cell row cell 3 left parenthesis 13 over 3 right parenthesis plus 2 left parenthesis negative 11 over 3 right parenthesis plus c end cell equals cell negative 13 end cell row cell 39 over 3 minus 22 over 3 plus c end cell equals cell negative 13 end cell row cell 17 over 3 plus c end cell equals cell negative 13 end cell row c equals cell negative 13 minus 17 over 3 end cell row c equals cell negative 39 over 3 minus 17 over 3 end cell row c equals cell negative 56 over 3 end cell end table 

Substitusikan nilai a comma space b comma space dan space c ke bentuk umum persamaan lingkaran

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus y squared plus a x plus b y plus c end cell equals 0 row cell x squared plus y squared plus left parenthesis 13 over 3 right parenthesis x plus left parenthesis negative 11 over 3 right parenthesis y plus left parenthesis negative 56 over 3 right parenthesis end cell equals 0 row cell x squared plus y squared plus 13 over 3 x minus 11 over 3 y minus 56 over 3 end cell equals 0 end table 

Jadi, persamaan lingkaran yang melalui titik left parenthesis 3 comma space 2 right parenthesis comma space left parenthesis negative 2 comma space 7 right parenthesis comma space dan space left parenthesis negative 4 comma space minus 3 right parenthesis adalah Error converting from MathML to accessible text. 

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

1

Iklan

Iklan

Pertanyaan serupa

Persamaan lingkaran yang melalui titik K ( 5 , 2 ) , L ( − 1 , 2 ) , dan M ( 3 , 6 ) adalah...

35

4.5

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia