A) ±16
B) ± 8
C) ± 1
D) ± 4
Correct Answer: D
Solution :
We know that, if the circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0,\]touches\[x-\]axis, then \[{{g}^{2}}=c\] Since,\[{{x}^{2}}+{{y}^{2}}+2ax+8y+16=0\]touches\[x-\]axis \[\therefore \] \[{{a}^{2}}=16\] \[\Rightarrow \] \[a=\pm 4\]You need to login to perform this action.
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