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For the two circles \mathrm{x^2+y^2=16}\mathrm{x^2+y^2=16}  and  \mathrm{x^2+y^2-2 y=0}, there is / are

Option: 1

one pair of common tangents

 


Option: 2

 two pairs of common tangents
 


Option: 3

three common tangents
 


Option: 4

no common tangents


Answers (1)

best_answer

\mathrm{S}_1: \mathrm{x}^2+\mathrm{y}^2-16=0
and, \mathrm{S_2: x^2+y^2-2 y=0}
Given circles are
Centre of \mathrm{S_1}  is \mathrm{C_1:(0,0)} and radius \mathrm{r_1=4}
Centre of \mathrm{S}_2 \text{ is }\mathrm{C}_2:(0,1)\text{ and radius }\mathrm{r}_2=1         \therefore \mathrm{C}_1 \mathrm{C}_2=\sqrt{0+1}=1

Since \left|\mathrm{C}_1 \mathrm{C}_2\right|<\left|\mathrm{r}_1 \mathrm{r}_2\right|, \therefore \mathrm{S}_2  is completely within \mathrm{S}_1  and hence there are no common tangents to the two circles.

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