Roboguru

Diketahui lingkaran  konsentris (sepusat) dengan lingkaran  dan melalui titik . Jika persamaan lingkaran , maka tentukan persamaan lingkaran .

Pertanyaan

Diketahui lingkaran begin mathsize 14px style straight L subscript 1 end style konsentris (sepusat) dengan lingkaran begin mathsize 14px style straight L subscript 2 end style dan melalui titik begin mathsize 14px style left parenthesis 2 comma space 8 right parenthesis end style. Jika persamaan lingkaran begin mathsize 14px style x squared plus y squared minus x plus 2 y minus 5 equals 0 end style, maka tentukan persamaan lingkaran undefined.

Pembahasan Soal:

Diketahui:

Persamaan lingkaran begin mathsize 14px style straight L subscript 2 end style adalah begin mathsize 14px style x squared plus y squared minus x plus 2 y minus 5 equals 0 end style
begin mathsize 14px style straight L subscript 1 end style melalui begin mathsize 14px style left parenthesis 2 comma space 8 right parenthesis end style

Mencari pusat:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight L subscript 2 end cell equals cell open parentheses negative straight A over 2 comma negative straight B over 2 close parentheses end cell row blank equals cell open parentheses negative open parentheses negative thin space 1 half close parentheses comma thin space open parentheses negative 2 over 2 close parentheses close parentheses end cell row blank equals cell open parentheses 1 half comma thin space minus 1 close parentheses end cell end table end style 

Sehingga pusat begin mathsize 14px style straight L subscript 1 equals straight L subscript 2 equals open parentheses 1 half comma space minus 1 close parentheses end style 

mencari persamaan begin mathsize 14px style straight L subscript 1 end style melalui begin mathsize 14px style left parenthesis 2 comma space 8 right parenthesis end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses straight x minus straight x subscript 0 close parentheses squared plus open parentheses straight y minus straight y subscript 0 close parentheses squared end cell equals cell straight r squared end cell row cell open parentheses straight x minus 1 half close parentheses squared plus open parentheses straight y minus open parentheses negative 1 close parentheses close parentheses squared end cell equals cell straight r squared thin space end cell row cell open parentheses straight x minus 1 half close parentheses squared plus open parentheses straight y plus 1 close parentheses squared end cell equals cell straight r squared thin space end cell row cell open parentheses 2 minus 1 half close parentheses squared plus open parentheses 8 plus 1 close parentheses squared end cell equals cell straight r squared thin space end cell row cell open parentheses 3 over 2 close parentheses squared plus 9 squared end cell equals cell straight r squared thin space end cell row cell 9 over 4 plus 81 end cell equals cell straight r squared end cell row cell fraction numerator 9 plus 324 over denominator 4 end fraction end cell equals cell straight r squared end cell row cell 333 over 4 end cell equals cell straight r squared end cell end table end style 

Sehingga didapat begin mathsize 14px style straight L subscript 1 end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x minus 1 half close parentheses squared plus open parentheses y plus 1 close parentheses squared end cell equals cell 333 over 4 end cell row cell thin space x squared minus x plus 1 fourth plus y squared plus 2 y plus 1 end cell equals cell 333 over 4 thin space end cell row cell x squared plus y squared minus x plus 2 y plus 1 fourth plus 1 end cell equals cell 333 over 4 thin space end cell row cell 4 x squared plus 4 y squared minus 4 x plus 8 y plus 1 plus 4 minus 333 end cell equals cell 0 thin space thin space end cell row cell 4 x squared plus 4 y squared minus 4 x plus 8 y minus 328 end cell equals cell 0 thin space thin space end cell end table end style 

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

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 15 Maret 2021

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