- Home
- Standard 11
- Mathematics
10-1.Circle and System of Circles
hard
વર્તુળો $x^2 +y^2 - 8x - 2y + 1 = 0$ અને $x^2 + y^2 + 6x + 8y = 0$ ને સામાન્ય સ્પર્શકોની સંખ્યા મેળવો.
A
એક
B
ચાર
C
બે
D
ત્રણ
(AIEEE-2012)
Solution
Given circles are
${x^2} + {y^2} – 8x + 2y + 1 = 0\,\,$
and ${x^2} + {y^2} + 6x + 8y + 1 = 0\,\,$
Their centres and radius are
${C_1}\left( {4,1} \right),{r_1} = \sqrt {16} = 4$
${C_2}\left( { – 3, – 4} \right),{r_1} = \sqrt {25} = 5$
Now, ${C_1}{C_2} = \sqrt {49 + 25} = \sqrt {74} $
${r_1} – {r_2} = – 1,{r_1} + {r_2} = 9$
Since, ${r_1} – {r_2} < {C_1}{C_2} < {r_1} + {r_2}$
$\therefore $ Number of common tangents $=2$
Standard 11
Mathematics
