The question is: does the distance between the point $(2,6)$ to the that line could be $5$?
is there a solution to the problem without computing?
i would glad to know.
thanks.
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The problem is wrongly posed as there are infinite lines through $(3,0),$ (one of them, btw, through $(2,6),$ , so without knowing what line exactly one cannot address the question. – DonAntonio Jun 06 '13 at 11:46
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what do you mean by "without computing"? – mau Jun 06 '13 at 11:47
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@mau - without algebraic solution, just by explaining in words. – bori12 Jun 06 '13 at 11:49
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my answer should then suffice. – mau Jun 06 '13 at 11:50
3 Answers
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Draw a circle with center $(2,6)$ and radius $5$
We need to show the existence of a tangent passing through $(3,0)$
As the distance between the points $(3,0),(2,6)$ is $\sqrt{(3-2)^2+(0-6)^2}>5,$ the point $(3,0)$ is outside the circle, we know there are two tangents
lab bhattacharjee
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thanks. btw, is there a problem with my question as DonAntonio mentioned? i asked if it could be and not the line itself – bori12 Jun 06 '13 at 12:05
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another question how the redius is 5 and the length of the Perpendicular is also 5? – bori12 Jun 06 '13 at 12:45
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@bori12, the perpendicular distance of a tangent from the centre of the circle = radius of the circle – lab bhattacharjee Jun 06 '13 at 14:32
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since the line $y=0$ has distance 6 from the point $(2,6)$ and there is a line with distance 0 (the one through the two points) and the distance is a continuous function, yes: there are (two) lines with distance 5. If you want to know which lines, that's another matter.
mau
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Yes, because the difference in $y$-coordinates is more than 5, so the distance is also more than 5 (I hope that's not too much computing!).
If the points would be $(3,0)$ and $(7,4)$, this simple argument doesn't work anymore and you need to start using Pythagoras.
yatima2975
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