A) 1
B) -1
C) \[\frac{3}{2}\]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
The equation of tangent at point \[(-2,\ -3)\] to the circle \[{{x}^{2}}+{{y}^{2}}+2x+4y+3=0\] is, \[-2x-3y+1(x-2)+2(y-3)+3=0\] \[\Rightarrow -2x-3y+x-2+2y-6+3=0\] \[\Rightarrow -x-y-5=0\Rightarrow x+y+5=0\] or\[y=-x-5\]; so, \[m=-1\] Hence, gradient of normal\[=\frac{-1}{-1}=1\].You need to login to perform this action.
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