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Keliling irisan lingkaranx2+y2−2x−6y+6=0 dan x2+y2−2x−2y−6=0 adalah ….

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Keliling irisan lingkaranbegin mathsize 14px style x squared plus y squared minus 2 x minus 6 y plus 6 equals 0 end style dan begin mathsize 14px style x squared plus y squared minus 2 x minus 2 y minus 6 equals 0 end style adalah ….

  1. begin mathsize 14px style fraction numerator cos to the power of negative 1 end exponent open parentheses 3 over 4 close parentheses over denominator 2 pi end fraction cross times 2 straight pi square root of 2 plus 2 pi end style 

  2. begin mathsize 14px style fraction numerator cos to the power of negative 1 end exponent open parentheses 3 over 4 close parentheses over denominator 2 pi end fraction cross times 4 straight pi square root of 2 plus 2 pi end style 

  3. begin mathsize 14px style fraction numerator cos to the power of negative 1 end exponent open parentheses 3 over 4 close parentheses over denominator 2 pi end fraction cross times 4 straight pi square root of 2 plus 4 pi end style 

  4. begin mathsize 14px style fraction numerator cos to the power of negative 1 end exponent open parentheses 3 over 4 close parentheses over denominator 4 pi end fraction cross times 2 straight pi square root of 2 plus 4 pi end style 

  5. begin mathsize 14px style fraction numerator cos to the power of negative 1 end exponent open parentheses 3 over 4 close parentheses over denominator 4 pi end fraction cross times 4 straight pi square root of 2 plus 4 pi end style 

H. Nufus

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Pembahasan

Pusat dan jaari-jari lingkaran begin mathsize 14px style x squared plus y squared minus 2 x minus 2 y minus 6 equals 0 end style yaitu

begin mathsize 14px style straight P subscript 1 open parentheses fraction numerator negative 2 over denominator negative 2 end fraction comma fraction numerator negative 2 over denominator negative 2 end fraction close parentheses equals straight P subscript 1 open parentheses 1 , 1 close parentheses straight r equals square root of 1 fourth open parentheses negative 2 close parentheses squared plus 1 fourth open parentheses negative 2 close parentheses squared plus 6 end root equals square root of 8 end style 

Pertama, tentukan titik potongan kedua lingkaran.

begin mathsize 14px style straight L subscript 1 blank colon blank x squared plus y squared minus 2 x minus 6 y plus 6 equals 0 bottom enclose straight L subscript 2 blank colon blank x squared plus y squared minus 2 x minus 2 y minus 6 equals 0 space minus end enclose space space space space space space space space space space space space space space space space space space space space space space space minus 4 y plus 12 equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space y equals 3 end style 

Substitusi nilai y=3 ke persamaan lingkaran 1 atau 2 untuk memperoleh nilai y.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus y squared minus 2 x minus 6 y plus 6 end cell equals 0 row cell x squared plus 3 squared minus 2 x minus 6 open parentheses 3 close parentheses plus 6 end cell equals 0 row cell x squared plus 9 minus 2 x minus 18 plus 6 end cell equals 0 row cell x squared minus 2 x minus 3 end cell equals 0 row cell open parentheses x plus 1 close parentheses open parentheses x minus 3 close parentheses end cell equals 0 end table end style 

Sehingga diperoleh x=-1 atau x=3.
Perpotongan kedua lingkaran yaitu pada koordinat A(-1,3)  dan B(3,3).
Maka panjang AB begin mathsize 14px style equals square root of open parentheses negative 1 minus 3 close parentheses squared plus open parentheses 3 minus 3 close parentheses squared end root equals 4 end style
Perhatikan gambar kedua lingkaran berikut ini.

 

Misalkan:

O adalah titik pusat lingkaran kedua.

Besar sudut AOB dapat dihitung dengan rumus aturan cos berikut ini.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos angle AOB end cell equals cell fraction numerator AO squared plus OB squared minus AB squared over denominator 2 times AO times OB end fraction end cell row blank equals cell fraction numerator open parentheses square root of 8 close parentheses squared plus open parentheses square root of 8 close parentheses squared minus 4 over denominator 2 open parentheses square root of 8 close parentheses open parentheses square root of 8 close parentheses end fraction end cell row blank equals cell 12 over 16 end cell row blank equals cell 3 over 4 end cell row cell angle AOB end cell equals cell cos to the power of negative 1 end exponent open parentheses 3 over 4 close parentheses end cell row blank blank blank end table end style 

Sehingga, panjang tali busur kecil AB adalah

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cos to the power of negative 1 end exponent open parentheses 3 over 4 close parentheses over denominator 2 pi end fraction cross times Keliling blank straight L subscript 2 end cell equals cell fraction numerator cos to the power of negative 1 end exponent open parentheses 3 over 4 close parentheses over denominator 2 pi end fraction cross times 2 straight pi square root of 8 end cell row blank equals cell fraction numerator cos to the power of negative 1 end exponent open parentheses 3 over 4 close parentheses over denominator 2 straight pi end fraction cross times 4 straight pi square root of 2 end cell end table end style 

Sedangkan panjang tali busur besar AB sama dengan setengah dari lingkaran begin mathsize 14px style L subscript 1 end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 1 half blank Keliling blank straight L subscript 1 end cell equals cell 1 half times 2 pi r end cell row blank equals cell pi r end cell row blank equals cell 2 pi end cell end table end style 

Jadi, keliling irisan kedua lingkaran tersebut adalah

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