Find the equations of the lines through $(2,0)$ and tangent to the circle $x^2+y^2=1$. I tried to solve this and I know the right answer but I just can't solve this. The right answer: $\sqrt{3}y=x-2$ or $\sqrt{3}y=2-x$.
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HINT:
The equation of any line passing through $(2,0)$ can be written as $$\frac{y-0}{x-2}=m\iff y=m(x-2)$$ where $m$ is the gradient or slope
Now replace the value of $y$ in $x^2+y^2=1$ to form a Quadratic Equation in $x$
For tangency, the roots of the equation must be same i..e, the discriminant must be zero
lab bhattacharjee
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HINT:
The equation of any line passing through $(2,0)$ can be written as $$\frac{y-0}{x-2}=m\iff mx-y-2m=0\ \ \ \ (1)$$ where $m$ is the gradient or slope
From this, the distance of a tangent from the center of the respective circle equals to the radius
Now can you calculate the distance of $(1)$ from the center$(0,0)$?
lab bhattacharjee
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@user146411, What's your Question? – lab bhattacharjee Apr 29 '14 at 14:46
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The distance of 1 from the center (0,0) is 1? – user146411 Apr 29 '14 at 15:06
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@user146411, yes – lab bhattacharjee Apr 29 '14 at 15:07
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@user146411, equate the radius($=1$) with the perpendicular distance – lab bhattacharjee Apr 29 '14 at 15:20
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1×0+1×0-1÷√1+1 ≫ -1÷√2 ≫ - √2÷2 I don't know what to do next – user146411 Apr 29 '14 at 15:33