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10-1.Circle and System of Circles
hard
વર્તૂળો ${x^2} + {y^2} = 4$ અને ${x^2} + {y^2} - 6x - 8y = 24$ ના સામાન્ય સ્પર્શકોની સંખ્યા મેળવો.
A
$0$
B
$1$
C
$3$
D
$4$
(IIT-1998)
Solution
(b) Circles ${S_1} \equiv {x^2} + {y^2} = {2^2}$,
${S_2} \equiv {(x – 3)^2} + {(y – 4)^2} = {7^2}$
$\therefore $ Centres ${C_1} = (0,\;0),$ ${C_2} = (3,\,4)$
and radii ${r_1} = 2,\;{r_2} = 7$
$\therefore \;{C_1}{C_2} = 5,$ ${r_2} – {r_1} = 5$
$i.e$., circles touch internally.
Hence there is only one common tangent.
Standard 11
Mathematics
