A) \[my=x\pm a\sqrt{1+{{m}^{2}}}\]
B) \[y=mx\pm \sqrt{1+{{m}^{2}}}\]
C) \[y=mx\pm a\sqrt{1+{{m}^{2}}}\]
D) None of these
Correct Answer: C
Solution :
Equation of tangent to the circle\[{{x}^{2}}+{{y}^{2}}={{a}^{2}},\]which is parallel to\[y=mx+c\]is \[y=mx\pm a\sqrt{1+{{m}^{2}}}\]You need to login to perform this action.
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