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10-1.Circle and System of Circles
hard
For the two circles $x^2 + y^2 = 16$ and $x^2 + y^2 -2y = 0,$ there is/are
A
one pair of common tangents
B
two pair of common tangents
C
three pair of common tangents
D
no common tangent
(JEE MAIN-2014)
Solution
Let, ${x^2} + {y^2} = 16$ or${x^2} + {y^2} = {4^2}$
radius of circle ${r_1} = 4\,$ center ${C_1}\left( {0,0} \right)$
we have ${x^2} + {y^2} – 2y = 0$
or ${x^2} + {\left( {y – 1} \right)^2} = {1^2}$
Rsdius $1$, center ${C_2}\left( {0,1} \right)$
$\left| {{C_1}{C_2}} \right| = 1$
$\left| {{r_2} – {r_1}} \right| = \left| {4 – 1} \right| = 3$
$\left| {{C_1}{C_2}} \right| > \left| {{r_2} – {r_1}} \right|$
no common tangents for these two circles.
Standard 11
Mathematics
