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Point $(1,2)$ relative to the circle $x^{2}+y^{2}+4 x-2 y-4=0$ is a/ an
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Verified Answer
The correct answer is:
exterior point
We put the co-ordinates of the given point in the given equation of circle $x^{2}+y^{2}+4 x-2 y-4=0$
$\operatorname{At}(1,2)$
$(1)^{2}+(2)^{2}+4(1)-2(2)-4$
$=1+4+4-4-4=1>0$
$\Rightarrow$ Point $(1,2)$ lies out side the circle i.e, an exterior point.
$\operatorname{At}(1,2)$
$(1)^{2}+(2)^{2}+4(1)-2(2)-4$
$=1+4+4-4-4=1>0$
$\Rightarrow$ Point $(1,2)$ lies out side the circle i.e, an exterior point.
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