JEE Main 2014 12 April Online
Exam JEE Main
Year 2014
Date 12 April
Mode Online
Subject Mathematics
Chapter Circle JEE Main 2014 · 12 April · Online · Mathematics · Circle
For the two circles x^2+y^2=16 and x^2+y^2-2 y=0 , there is/are
For the two circles
x 2 + y 2 = 16 x^2+y^2=16 x 2 + y 2 = 16 and
x 2 + y 2 − 2 y = 0 x^2+y^2-2 y=0 x 2 + y 2 − 2 y = 0 , there is/are
Options A one pair of common tangents B two pair of common tangents C three pair of common tangents D no common tangent Correct answer D. no common tangent
Step-by-step solution
Let,
x 2 + y 2 = 16 x^2+y^2=16 x 2 + y 2 = 16 or
x 2 + y 2 = 4 2 x^2+y^2=4^2 x 2 + y 2 = 4 2 radius of circle
r 1 = 4 r_1=4 r 1 = 4 , centre
C 1 ( 0 , 0 ) \mathrm{C}_1(0,0) C 1 ( 0 , 0 ) we have,
x 2 + y 2 − 2 y = 0 x^2+y^2-2 y=0 x 2 + y 2 − 2 y = 0 ⇒ x 2 + ( y 2 − 2 y + 1 ) − 1 = 0 or x 2 + ( y − 1 ) 2 = 1 2 \begin{aligned}
\Rightarrow & x^2+\left(y^2-2 y+1\right)-1=0 \text{ or } x^2+(y-1)^2 \\
=& 1^2
\end{aligned} ⇒ = x 2 + ( y 2 − 2 y + 1 ) − 1 = 0 or x 2 + ( y − 1 ) 2 1 2 R a d i u s 1 , c e n t r e C 2 ( 0 , 1 ) Radius 1, centre \mathrm{C}_2(0,1) R a d i u s 1 , ce n t re C 2 ( 0 , 1 ) ∣ C 1 C 2 ∣ = 1 ∣ r 2 − r 1 ∣ = ∣ 4 − 1 ∣ = 3 ∣ C 1 C 2 ∣ < ∣ r 2 − r 1 ∣ \begin{aligned}
&\left|C_1 C_2\right|=1 \\
&\left|r_2-r_1\right|=|^{4}-1|=3 \\
&\left|C_1 C_2\right| < \left|r_2-r_1\right|
\end{aligned} ∣ C 1 C 2 ∣ = 1 ∣ r 2 − r 1 ∣ = ∣ 4 − 1∣ = 3 ∣ C 1 C 2 ∣ < ∣ r 2 − r 1 ∣ n o c o m m o n t a n g e n t s f o r t h e s e t w o c i r c l e s . no common tangents for these two circles. n oco mm o n t an g e n t s f or t h ese tw oc i rc l es .
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