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Two tangents of the circle $x^2+y^2=8$ at $A$ and $B$ meet at $P =(-4,0)$ then find the area of quadrilateral $PAOB$.

Here, $O$ is the origin.

Distance of $P$ from $O$ is $4$ unit.

$$OA=OB=2\sqrt 2$$

Therefore $$OP^2=OA^2+AP^2$$ $$AP=2\sqrt 2$$

Hence $$\Delta POA=4$$ $$ar(PAOB)=8$$

This was a true of false question. I modified it a bit. Basically, the answer says the area is not 8sq unit. What am I doing wrong.

There is another sub question to this

Is the area of quadrilateral formed by the tangents from an external point with length of tangent 2r?

From the way I have solved, I am inclined to say no, but since it says my answer is wrong, please help me verify this.

lioness99a
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Aditya
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  • Why 8 isn't the answer? – Jo Jomax Feb 05 '20 at 16:42
  • I think your computation of the area is correct. Maybe there is something wrong in the solutions? – Unknown Feb 05 '20 at 16:58
  • @Unknown that’s what I thought, but needed to get it checked anyway. Also, the second statement remains untrue too right? – Aditya Feb 05 '20 at 17:08
  • @Aditya i am not quite sure, if I understand that subquestion right, but assuming that the "length of tangent" means the distance from $P$ to $A$ and $r$ is the radius of the circle I would say so, yes. – Unknown Feb 05 '20 at 17:28
  • @Unknown 'Yes' means the statement is false? – Aditya Feb 05 '20 at 17:57
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    @Aditya if my understanding of the question is correct, then I would say the statement is false. Sorry for confusing you. – Unknown Feb 05 '20 at 18:00

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