Pertanyaan

(SNMPTN 2011) Lingkaran dengan pusat (2, 3) dan menyinggung garis y = 2 x adalah ….

(SNMPTN 2011)

Lingkaran dengan pusat (2, 3) dan menyinggung garis $y = 2 x$ adalah ….

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Klaim

I. Roy

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Pembahasan

$begin mathsize 14px style J a r i minus j a r i equals j a r a k space t i t i k space left parenthesis 2 comma space 3 right parenthesis space k e space g a r i s space y equals 2 x left right arrow 2 x minus y equals 0 r equals open vertical bar fraction numerator 2.2 plus 3. left parenthesis negative 1 right parenthesis plus 0 over denominator square root of 2 squared plus left parenthesis negative 1 right parenthesis squared end root end fraction close vertical bar space r equals open vertical bar fraction numerator 1 over denominator square root of 5 end fraction close vertical bar space r equals fraction numerator 1 over denominator square root of 5 end fraction space J a d i comma space p e r s a m a a n space l i n g k a r a n space d e n g a n space p u s a t space left parenthesis 2 comma space 3 right parenthesis d a n space j a r i minus j a r i space r equals fraction numerator 1 over denominator square root of 5 end fraction space a d a l a h space colon space left parenthesis x minus h right parenthesis squared plus left parenthesis y minus k right parenthesis squared equals r squared space left parenthesis x minus 2 right parenthesis squared plus left parenthesis y minus 3 right parenthesis squared equals open parentheses fraction numerator 1 over denominator square root of 5 end fraction close parentheses squared space space space x squared minus 4 x plus 4 plus y squared minus 6 y plus 9 equals 1 fifth space x squared plus y squared minus 4 x minus 6 y plus 13 equals 1 fifth left parenthesis k e d u a space r u a s space d i k a l i k a n space d e n g a n space 5 right parenthesis space S e h i n g g a space colon space 5 x squared plus 5 y squared minus 20 x minus 30 y plus 65 equals 1 5 x squared plus 5 y squared minus 20 x minus 30 y plus 64 equals 0 end style$