Roboguru

Tentukan persamaan lingkaran yang melalui titik potong lingkaran  dan  serta melalui titik

Pertanyaan

Tentukan persamaan lingkaran yang melalui titik potong lingkaran L subscript 1 equals x squared plus y squared plus 2 x plus 2 y minus 2 equals 0 dan L subscript 2 equals x squared plus y squared plus 4 x minus 8 y plus 4 equals 0 comma serta melalui titik left parenthesis 2 comma negative 1 right parenthesis. 

Pembahasan Soal:

Langkah pertama: Substitusikan titik left parenthesis 2 comma negative 1 right parenthesis ke persamaan L subscript 1 dan L subscript 2 sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell L subscript 1 end cell equals cell x squared plus y squared plus 2 x plus 2 y minus 2 equals 0 end cell row blank equals cell 2 squared plus left parenthesis negative 1 right parenthesis squared plus 2 times 2 plus 2 times left parenthesis negative 1 right parenthesis minus 2 end cell row blank equals cell 4 plus 1 plus 4 minus 2 minus 2 end cell row cell L subscript 1 end cell equals 5 end table  

table attributes columnalign right center left columnspacing 0px end attributes row cell L subscript 2 end cell equals cell x squared plus y squared plus 4 x minus 8 y plus 4 equals 0 end cell row blank equals cell 2 squared plus left parenthesis negative 1 right parenthesis squared plus 4 times 2 minus 8 times negative 1 plus 4 end cell row blank equals cell 4 plus 1 plus 8 plus 8 plus 4 end cell row cell L subscript 2 end cell equals 25 end table 

Langkah kedua: Misalkan persamaan lingkaran yang dicari adalah left parenthesis L subscript 3 right parenthesis, maka berlaku:

table attributes columnalign right center left columnspacing 0px end attributes row cell L subscript 3 end cell equals cell L subscript 1 plus lambda times L subscript 2 equals 0 end cell row cell L subscript 3 end cell equals cell open parentheses x squared plus y squared plus 2 x plus 2 y minus 2 close parentheses plus lambda times open parentheses x squared plus y squared plus 4 x minus 8 y plus 4 close parentheses equals 0 end cell end table      

Substitusikan nilai L subscript 1 dan L subscript 2  ke persamaan Error converting from MathML to accessible text. menentukan nilai lambda sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell L subscript 1 plus lambda times L subscript 2 end cell equals 0 row cell 5 plus 25 lambda end cell equals 0 row lambda equals cell negative 5 over 25 end cell row lambda equals cell negative 1 fifth end cell end table 

 Langkah ketiga: Akan ditentukan persamaan lingkaran left parenthesis L subscript 3 right parenthesis dengan cara substitusikan nilai lambda ke persamaan left parenthesis L subscript 3 right parenthesis di atas sebagai berikut.

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x squared plus y squared plus 2 x plus 2 y minus 2 close parentheses plus lambda times open parentheses x squared plus y squared plus 4 x minus 8 y plus 4 close parentheses end cell equals 0 row cell open parentheses x squared plus y squared plus 2 x plus 2 y minus 2 close parentheses minus 1 fifth open parentheses x squared plus y squared plus 4 x minus 8 y plus 4 close parentheses end cell equals 0 row cell x squared plus y squared plus 2 x plus 2 y minus 2 minus 1 fifth x squared minus 1 fifth y squared minus 4 over 5 x plus 8 over 5 y minus 4 over 5 end cell equals 0 row cell fraction numerator 5 x squared minus x squared over denominator 5 end fraction plus fraction numerator 5 y squared minus y squared over denominator 5 end fraction plus fraction numerator 10 x minus 4 x over denominator 5 end fraction plus fraction numerator 10 y plus 8 y over denominator 5 end fraction plus fraction numerator negative 10 minus 4 over denominator 5 end fraction end cell equals 0 row cell 4 over 5 x squared plus 4 over 5 y squared plus 6 over 5 x plus 18 over 5 y minus 14 over 5 end cell equals cell 0 space open parentheses cross times 5 close parentheses end cell row cell 4 x squared plus 4 y squared plus 6 x plus 18 y minus 14 end cell equals cell 0 left parenthesis divided by 2 right parenthesis end cell row cell 2 x squared plus 2 y squared plus 3 x plus 9 y minus 7 end cell equals 0 end table end style    

Dengan demikian, persamaan lingkaran pada soal tersebut adalah 2 x squared plus 2 y squared plus 3 x plus 9 y minus 7 equals 0. 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Sibuea

Mahasiswa/Alumni Universitas Riau

Terakhir diupdate 04 Juni 2021

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Pertanyaan yang serupa

Diketahui 2 titik dengan koordinat  dan . Jika KM merupakan diameter lingkaran, maka persamaan lingkaran tersebut adalah...

Pembahasan Soal:

Persamaan umum lingkaran

open parentheses x minus a close parentheses squared plus open parentheses y minus b close parentheses squared equals r squared 

Mencari diameter dan jari-jari lingkaran

table attributes columnalign right center left columnspacing 0px end attributes row d equals cell K M end cell row blank equals cell square root of open parentheses x subscript 1 minus x subscript 2 close parentheses squared plus open parentheses y subscript 1 minus y subscript 2 close parentheses squared end root end cell row blank equals cell square root of open parentheses negative 3 minus 3 close parentheses squared plus open parentheses 4 minus 8 close parentheses squared end root end cell row blank equals cell square root of open parentheses negative 6 close parentheses squared plus open parentheses negative 4 close parentheses squared end root end cell row blank equals cell square root of 36 plus 16 end root end cell row blank equals cell square root of 52 end cell row blank equals cell square root of 4 cross times 13 end root end cell row blank equals cell 2 square root of 13 space cm end cell row r equals cell 1 half d end cell row blank equals cell 1 half cross times 2 square root of 13 end cell row blank equals cell square root of 13 space cm end cell end table 

Mencari titik pusat lingkaran yaitu titik tengah KM

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a comma b close parentheses end cell equals cell open parentheses fraction numerator x subscript 1 plus x subscript 2 over denominator 2 end fraction comma fraction numerator y subscript 1 plus y subscript 2 over denominator 2 end fraction close parentheses end cell row blank equals cell open parentheses fraction numerator negative 3 plus 3 over denominator 2 end fraction comma fraction numerator 4 plus 8 over denominator 2 end fraction close parentheses end cell row blank equals cell open parentheses 0 comma 6 close parentheses end cell end table  

Menentukan persamaan lingkaran

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 0 close parentheses squared plus open parentheses y minus 6 close parentheses squared end cell equals cell open parentheses square root of 13 close parentheses squared end cell row cell x squared plus y squared minus 12 y plus 36 end cell equals 13 row cell x squared plus y squared minus 12 y plus 23 end cell equals 0 end table 

Jadi, persamaan lingkarannya adalah Error converting from MathML to accessible text..

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Roboguru

Tentukan persamaan umum lingkaran yang melalui titik

Pembahasan Soal:

Persamaan umum lingkaran adalah begin mathsize 14px style straight x squared plus straight y squared plus Ax plus By plus straight C equals 0 end style.

Substitusikan begin mathsize 14px style straight P open parentheses 6 comma negative 2 close parentheses end style ke dalam persamaan begin mathsize 14px style straight x squared plus straight y squared plus Ax plus By plus straight C equals 0 end style.

x2+y2+Ax+By+C62+(2)2+A(6)+B(2)+C36+4+6A2B+C6A2B+C====00040...(i) 

Substitusikan begin mathsize 14px style straight Q open parentheses negative 3 comma negative 5 close parentheses end style ke dalam persamaan begin mathsize 14px style straight x squared plus straight y squared plus Ax plus By plus straight C equals 0 end style.

x2+y2+Ax+By+C(3)2+(5)2+A(3)+B(5)+C9+253A5B+C3A5B+C====00034...(ii) 

Substitusikan begin mathsize 14px style straight R open parentheses 1 comma 3 close parentheses end style ke dalam persamaan begin mathsize 14px style straight x squared plus straight y squared plus Ax plus By plus straight C equals 0 end style.

x2+y2+Ax+By+C12+(3)2+A(1)+B(3)+C1+9+A+3B+CA+3B+C====00010...(iii) 

Eliminasi persamaan (i) dan (ii)

6A2B+C=403A5B+C=349A+3B=6...(iv)

Eliminasi persamaan (i) dan (iii)

begin mathsize 14px style 6 straight A minus 2 straight B plus straight C equals negative 40 space space straight A plus 3 straight B plus straight C equals negative 10 space minus stack 5 straight A minus 5 straight B space space space space space space equals negative 30 space space space space with bar on top... space left parenthesis straight v right parenthesis end style

Eliminasi persamaan (iv) dan (v)

5A5B=309A+3B=6×3∣×5∣15A15B=9045A+15B=30+60A=120A=2 

Substitusi nilai A = 2 pada salah satu persamaan, kita ambil persamaan v diperoleh:

5A5B5(2)5B105B5BB=====303030204 

Substitusi nilai A = 2 dan B = 4 pada salah satu persamaan, kita ambil persamaan iii diperoleh:

A+3B+C(2)+3(4)+C2+12+C10+CC=====1010101020 

Dengan demikian, persamaan umum lingkaran tersebut adalah begin mathsize 14px style straight x squared plus straight y squared minus 2 straight x plus 4 straight y minus 20 equals 0 end style.

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Roboguru

Lingkaran yang melalui titik-titik   dan  berjari-jari

Pembahasan Soal:

Dengan menggunakan rumus persamaan lingkaran dengan pusat open parentheses a comma b close parentheses dan jari-jari open parentheses r close parentheses, maka berlaku:

x squared plus y squared plus A x plus B y plus C equals 0 

Maka,

Untuk titik open parentheses 4 comma 2 close parentheses, diperoleh:

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 4 squared plus 2 squared plus A times 4 plus B times 2 plus C end cell equals 0 row cell 16 plus 4 plus 4 A plus 2 B plus C end cell equals 0 row cell 4 A plus 2 B plus C end cell equals cell negative 20 space space rightwards arrow text pers. end text space open parentheses 1 close parentheses end cell end table 

Untuk titik open parentheses 1 comma 3 close parentheses, diperoleh:

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 1 squared plus 3 squared plus A times 1 plus B times 3 plus C end cell equals 0 row cell 1 plus 9 plus A plus 3 B plus C end cell equals 0 row cell A plus 3 B plus C end cell equals cell negative 10 space space rightwards arrow text pers. end text space open parentheses 2 close parentheses end cell end table 

Untuk titik open parentheses negative 3 comma negative 5 close parentheses, diperoleh:

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 open parentheses negative 3 close parentheses squared plus open parentheses negative 5 close parentheses squared plus A times negative 3 plus B times negative 5 plus C end cell equals 0 row cell 9 plus 25 minus 3 A minus 5 B plus C end cell equals 0 row cell negative 3 A minus 5 B plus C end cell equals cell negative 34 space space rightwards arrow text pers. end text space open parentheses 3 close parentheses end cell end table 

Selanjutnya, gunakan metode eliminasi untuk mengeliminasi variabel open parentheses C close parentheses dari persamaan open parentheses 1 close parentheses space text dan end text space open parentheses 2 close parentheses sehingga diperoleh:

stack attributes charalign center stackalign right end attributes table row cell 4 A end cell plus cell 2 B end cell plus C equals cell negative 20 end cell end table table row cell space space A end cell plus cell 3 B end cell plus C equals cell negative 10 end cell end table horizontal line table row cell space space space space space space space space space table row cell space space 3 A end cell minus B equals cell negative 10 end cell end table end cell end table end stack negative space rightwards arrow text pers. end text space open parentheses 4 close parentheses  

Kemudian, eliminasi variabel open parentheses C close parentheses dari persamaan open parentheses 1 close parentheses space text dan end text space open parentheses 3 close parentheses sehingga diperoleh:

stack attributes charalign center stackalign right end attributes table row cell space space space 4 A end cell plus cell 2 B end cell plus C equals cell negative 20 end cell end table table row cell negative 3 A end cell minus cell 5 B end cell plus C equals cell negative 34 end cell end table horizontal line table row cell space space space space space space space space space table row cell 7 A end cell plus cell 7 B end cell equals 14 end table end cell row cell space space space space space space space space space space space table row A plus B equals 2 end table end cell end table end stack negative space space open parentheses colon space 2 close parentheses space rightwards arrow text pers. end text space open parentheses 5 close parentheses

Dari persamaan open parentheses 5 close parentheses diperoleh A equals 2 minus B. Substitusikan A equals 2 minus B ke persamaan open parentheses 4 close parentheses, maka akan diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 A minus B end cell equals cell negative 10 end cell row cell 3 open parentheses 2 minus B close parentheses minus B end cell equals cell negative 10 end cell row cell 6 minus 3 B minus B end cell equals cell negative 10 end cell row cell negative 4 B end cell equals cell negative 10 minus 6 end cell row B equals cell fraction numerator negative 16 over denominator negative 4 end fraction end cell row B equals 4 end table 

Karena nilai B equals 4, maka nilai A equals 2 minus B equals 2 minus 4 equals negative 2. Kemudian, substitusikan nilai A space text dan end text space B ke persamaan open parentheses 2 close parentheses untuk memperoleh nilai C sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell A plus 3 B plus C end cell equals cell negative 10 end cell row cell negative 2 plus 3 times 4 plus C end cell equals cell negative 10 end cell row cell negative 2 plus 12 plus C end cell equals cell negative 10 end cell row C equals cell negative 10 minus 10 end cell row C equals cell negative 20 end cell end table

Selanjutnya, substitusikan semua nilai yang sudah diperoleh ke rumus persamaan lingkaran sebagai berikut.

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 minus 2 x plus 4 y minus 20 end cell equals 0 row cell x squared minus 2 x plus y squared plus 4 y end cell equals 20 row cell x squared minus 2 x plus 1 plus y squared plus 4 y plus 4 end cell equals cell 20 plus 1 plus 4 end cell row cell open parentheses x minus 1 close parentheses squared plus open parentheses y plus 2 close parentheses squared end cell equals 25 row cell open parentheses x minus 1 close parentheses squared plus open parentheses y plus 2 close parentheses squared end cell equals cell 5 squared space rightwards arrow r equals 5 end cell end table 

Dengan demikian, jari-jari lingkaran yang melalui titik-titik open parentheses 4 comma 2 close parentheses comma open parentheses 1 comma 3 close parentheses dan open parentheses negative 3 comma negative 5 close parentheses adalah 5. 

0

Roboguru

Tentukan pusat, jari-jari, dan persamaan lingkaran yang melalui tiga titik , , dan

Pembahasan Soal:

Bentuk umum persamaan lingkaran, yaitu

x squared plus y squared plus A x plus B y plus C equals 0 dengan

text P end text open parentheses negative 1 half A comma space minus 1 half B close parentheses dan r equals square root of open parentheses negative 1 half A close parentheses squared plus open parentheses negative 1 half B close parentheses squared minus C end root 

Melalui titik text A end text open parentheses 3 comma space 1 close parentheses diperoleh persamaan 1 berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 squared plus 1 squared plus A times 3 plus B times 1 plus C end cell equals 0 row cell 9 plus 1 plus 3 A plus B plus C end cell equals 0 row cell 3 A plus B plus C end cell equals cell negative 10 end cell end table

Melalui titik text B end text open parentheses negative 2 comma space 6 close parentheses diperoleh persamaan 2 berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative 2 close parentheses squared plus 6 squared plus A open parentheses negative 2 close parentheses plus B times 6 plus C end cell equals 0 row cell 4 plus 36 minus 2 A plus 6 B plus C end cell equals 0 row cell negative 2 A plus 6 B plus C end cell equals cell negative 40 end cell end table

Melalui titik text C end text open parentheses negative 5 comma space minus 3 close parentheses diperoleh persamaan 3 berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative 5 close parentheses squared plus open parentheses negative 3 close parentheses squared plus A open parentheses negative 5 close parentheses plus B open parentheses negative 3 close parentheses plus C end cell equals 0 row cell 25 plus 9 minus 5 A minus 3 B plus C end cell equals 0 row cell negative 5 A minus 3 B plus C end cell equals cell negative 34 end cell end table

Dari persamaan 1 dan 2 diperoleh persamaan 4 berikut.

table row cell 3 A plus B plus C end cell equals cell negative 10 space space space space end cell row cell negative 2 A plus 6 B plus C end cell equals cell negative 40 space minus end cell row cell 5 A minus 5 B end cell equals 30 row cell A minus B end cell equals 6 end table

Dari persamaan 1 dan 3 diperoleh persamaan 5 berikut.

table row cell 3 A plus B plus C end cell equals cell negative 10 space space space space space end cell row cell negative 5 A minus 3 B plus C end cell equals cell negative 34 space minus end cell row cell space 8 A plus 4 B end cell equals 24 row cell 2 A plus B end cell equals 6 end table

Dari persamaan 4 dan 5 diperoleh

table row cell A minus B end cell equals cell 6 space space space space end cell row cell 2 A plus B end cell equals cell 6 space plus end cell row cell 3 A end cell equals 12 row A equals 4 end table

Substitusi A equals 4 ke persamaan 4 diperoleh nilai B berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell A minus B end cell equals 6 row cell 4 minus B end cell equals 6 row cell negative B end cell equals 2 row B equals cell negative 2 end cell end table

Substitusi A equals 4 dan B equals negative 2 ke persamaan 1 sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 A plus B plus C end cell equals cell negative 10 end cell row cell 3 times 4 plus open parentheses negative 2 close parentheses plus C end cell equals cell negative 10 end cell row cell 12 minus 2 plus C end cell equals cell negative 10 end cell row cell 10 plus C end cell equals cell negative 10 end cell row C equals cell negative 20 end cell end table

Persamaan lingkaran tersebut adalah x squared plus y squared plus 4 x minus 2 y minus 20 equals 0 

Pusat lingkaran, yaitu 

table attributes columnalign right center left columnspacing 0px end attributes row cell P open parentheses negative 1 half A comma space minus 1 half B close parentheses end cell equals cell P open parentheses negative 1 half times 4 comma space minus 1 half times open parentheses negative 2 close parentheses close parentheses end cell row blank equals cell P open parentheses negative 2 comma space 1 close parentheses end cell end table

Jari-jari lingkaran:

table attributes columnalign right center left columnspacing 0px end attributes row r equals cell square root of open parentheses negative 1 half A close parentheses squared plus open parentheses negative 1 half B close parentheses squared minus C end root end cell row blank equals cell square root of open parentheses negative 2 close parentheses squared plus 1 squared minus open parentheses negative 20 close parentheses end root end cell row blank equals cell square root of 4 plus 1 plus 20 end root end cell row blank equals cell square root of 25 end cell row blank equals 5 end table

Dengan demikian, persamaan lingkaran tersebut x squared plus y squared plus 4 x minus 2 y minus 20 equals 0 dengan pusat text P end text open parentheses negative 2 comma space 1 close parentheses dan jari-jari 5 

0

Roboguru

Tentukan persamaan lingkaran dengan pusat : b. melalui titik  .

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. 

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