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Tentukan persamaan lingkaran dengan pusat : b. melalui titik  .

Pertanyaan

Tentukan persamaan lingkaran dengan pusat begin mathsize 14px style straight O end style:

b. 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 .

Pembahasan Soal:

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. 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 27 Juli 2021

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