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Persamaan lingkaran yang berdiameter AB dengan A(−a,b) dan B(a,−b) adalah ....

Pertanyaan

Persamaan lingkaran yang berdiameter AB dengan straight A left parenthesis negative straight a comma straight b right parenthesis dan straight B left parenthesis straight a comma negative straight b right parenthesis adalah ....

  1. begin mathsize 14px style left parenthesis x squared minus a squared right parenthesis plus left parenthesis y squared minus b squared right parenthesis equals 0 end style  

  2. begin mathsize 14px style left parenthesis x squared plus a squared right parenthesis plus left parenthesis y squared minus b squared right parenthesis equals 0 end style  

  3. begin mathsize 14px style left parenthesis x squared minus a squared right parenthesis plus left parenthesis y squared plus b squared right parenthesis equals 0 end style 

  4. begin mathsize 14px style left parenthesis x squared plus a squared right parenthesis plus left parenthesis y squared plus b squared right parenthesis equals 0 end style 

  5. begin mathsize 14px style left parenthesis x minus a right parenthesis squared plus left parenthesis y minus b right parenthesis squared equals 0 end style 

G. Albiah

Master Teacher

Mahasiswa/Alumni Universitas Galuh Ciamis

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah A.

Pembahasan

Ingat!

  • Titik pusat lingkaran yaitu begin mathsize 14px style left parenthesis x subscript p comma y subscript p right parenthesis end style dapat ditentukan menggunakan rumus begin mathsize 14px style 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 style.
  • Jari-jari lingkaran dapat ditentukan dengan rumus begin mathsize 14px style r equals 1 half square root of open parentheses x subscript 2 minus x subscript 1 close parentheses squared plus open parentheses y subscript 2 minus y subscript 1 close parentheses squared end root end style.
  • Persamaan lingkaran dapat ditentukan dengan rumus begin mathsize 14px style open parentheses x minus x subscript p close parentheses squared plus open parentheses y minus y subscript p close parentheses squared equals r squared end style.

Diketahui,

  • Diameter lingkaran memiliki ujung pada titik  straight A left parenthesis negative straight a comma straight b right parenthesis dan straight B left parenthesis straight a comma negative straight b right parenthesis

Ditanyakan,

  • Persamaan lingkaran yang berdiameter AB dengan straight A left parenthesis negative straight a comma straight b right parenthesis space dan space straight B left parenthesis straight a comma negative straight b right parenthesis

- Menentukan titik pusat lingkaran:

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis x subscript p comma y subscript p right parenthesis 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 cell left parenthesis x subscript p comma y subscript p right parenthesis end cell equals cell open parentheses fraction numerator negative a plus a over denominator 2 end fraction comma fraction numerator b plus open parentheses negative b close parentheses over denominator 2 end fraction close parentheses end cell row cell left parenthesis x subscript p comma y subscript p right parenthesis end cell equals cell open parentheses fraction numerator a minus a over denominator 2 end fraction comma fraction numerator b minus b over denominator 2 end fraction close parentheses end cell row cell left parenthesis x subscript p comma y subscript p right parenthesis end cell equals cell open parentheses 0 over 2 comma 0 over 2 close parentheses end cell row cell left parenthesis x subscript p comma y subscript p right parenthesis end cell equals cell left parenthesis 0 , 0 right parenthesis end cell end table  

- Menentukan jari-jari lingkaran:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row r equals cell 1 half square root of open parentheses x subscript 2 minus x subscript 1 close parentheses squared plus open parentheses y subscript 2 minus y subscript 1 close parentheses squared end root end cell row r equals cell 1 half square root of open parentheses a minus left parenthesis negative a right parenthesis close parentheses squared plus open parentheses negative b minus b close parentheses squared end root end cell row r equals cell 1 half square root of open parentheses 2 a close parentheses squared plus open parentheses negative 2 b close parentheses squared end root end cell row r equals cell 1 half square root of 4 a squared plus 4 b squared end root end cell row r equals cell 1 half square root of 4 left parenthesis a squared plus b squared right parenthesis end root end cell row r equals cell 1 half open parentheses 2 close parentheses square root of a squared plus b squared end root end cell row r equals cell square root of a squared plus b squared end root end cell end table end style 

- Menentukan persamaan lingkaran:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x minus x subscript p close parentheses squared plus open parentheses y minus y subscript p 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 0 close parentheses squared end cell equals cell open parentheses square root of a squared plus b squared end root close parentheses squared end cell row cell x squared plus y squared end cell equals cell a squared plus b squared end cell row cell x squared plus y squared minus a squared minus b squared end cell equals 0 row cell x squared minus a squared plus y squared minus b squared end cell equals 0 row cell left parenthesis x squared minus a squared right parenthesis plus left parenthesis y squared minus b squared right parenthesis end cell equals 0 end table end style 

Jadi, persamaan lingkaran yang berdiameter AB dengan  straight A left parenthesis negative straight a comma straight b right parenthesis dan straight B left parenthesis straight a comma negative straight b right parenthesisadalah begin mathsize 14px style left parenthesis x squared minus a squared right parenthesis plus left parenthesis y squared minus b squared right parenthesis equals 0 end style  .

 

Oleh karena itu, jawaban yang tepat adalah A.

3rb+

4.7 (9 rating)

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