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
Solution :
Let, \[{{x}^{2}}+{{y}^{2}}=16\]or\[{{x}^{2}}+{{y}^{2}}={{4}^{2}}\]radius of circle \[{{r}_{1}}=4,\]centre \[{{C}_{1}}(0,0)\]we have, \[{{x}^{2}}+{{y}^{2}}-2y=0\] \[\Rightarrow \] \[{{x}^{2}}+({{y}^{2}}-2y+1)-1=0\] or\[{{x}^{2}}+{{(y-1)}^{2}}={{1}^{2}}\] Radius 1, centre \[{{C}_{2}}(0,1)\] \[|{{C}_{1}}{{C}_{2}}|=1\] \[|{{r}_{1}}-{{r}_{2}}|=|4-1|=3\] \[|{{C}_{1}}{{C}_{2}}|<|{{r}_{2}}-{{r}_{1}}|\] no common tangents for these two circles.
You need to login to perform this action.
You will be redirected in
3 sec
