Find, without differentiation, the equation of the straight lines which pass through the point $\displaystyle (4,0)$ and are tangential to the circle, $\displaystyle (x+1)^2+y^2=4$

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- Feb 25th 2011, 03:57 AMPunchFind the equation of the straight line
Find, without differentiation, the equation of the straight lines which pass through the point $\displaystyle (4,0)$ and are tangential to the circle, $\displaystyle (x+1)^2+y^2=4$

- Feb 25th 2011, 04:09 AMemakarov
If you draw a picture, there is a right triangle with two known sides. From there you can find angles, etc.

In general, you are supposed to show some effort. - Feb 25th 2011, 04:22 AMPunch
Sorry what I did was...

Circle center $\displaystyle = (-1,0)$

Radius=2

Let the unknown point of intersection be $\displaystyle (x,y)$

Length from circle center to $\displaystyle (4,0) = 5$

Since radius=2, the length of the line by pythagoras theorem$\displaystyle =\sqrt{21}$

$\displaystyle \sqrt{(x-4)^2+(y-0)^2}=\sqrt{21}$

$\displaystyle x^2-8x+16+y^2=21$

$\displaystyle x^2+y^2-8x-5=0$

$\displaystyle x^2+y^2=8x+5$

then, gradient of the radius from circle center to point of intersection $\displaystyle = \frac{1}{gradient of tangential line}$

$\displaystyle \frac{y-0}{x+1}=\frac{1}{\frac{y-0}{x-4}}$

$\displaystyle \frac{y}{x+1}=\frac{x-4}{y}$

$\displaystyle y^2=(x-4)(x+1)$

$\displaystyle y^2=x^2-3x-4$

sub $\displaystyle y^2=x^2-3x-4,$

$\displaystyle x^2+(x^2-3x-4)=8x+5$

$\displaystyle 2x^2-11x+1=0$

With the value of x found, i can find y but the equation of the line i got is different from the answer - Feb 25th 2011, 04:41 AMemakarov
I would do the following. The tangent of the angle at (4,0) is $\displaystyle 2/\sqrt{21}$. Therefore, the equations of the two lines are $\displaystyle y=\pm(x-4)\cdot2/\sqrt{21}$.

- Feb 25th 2011, 05:10 AMbjhopperfind equations of tangents
Hi punch,

After defining one of the tangency triangles you need to find the altitude on the hypothenuse and the segments it creats on the hypot.A slope of the tangent will then be definable.Next use point slope formula to find the lines.

bjh - Feb 25th 2011, 11:32 PMPunch
- Feb 26th 2011, 01:10 AMemakarov
https://lh4.googleusercontent.com/_S...00/tangent.png

Since $\displaystyle \angle ABC=90^\circ$, $\displaystyle \tan\angle BCA=AB/BC=2/\sqrt{21}$. Therefore, the slope of BC is $\displaystyle -2/\sqrt{21}$. Does this answer your question? - Feb 26th 2011, 02:57 AMPunch
- Feb 26th 2011, 03:21 AMbjhopper
Hi,

Emarkarov;s equations are correct but the question is " Is this solution in a geometry problem valid ? "

bjh - Feb 26th 2011, 05:32 AMemakarov
https://lh6.googleusercontent.com/_S...0/tangent1.png

I think, by definition, the gradient of line BC is $\displaystyle \tan\alpha$. (Please correct me if I am wrong.) Taken direction into account, $\displaystyle \alpha<0$. Also, $\displaystyle \angle ACB=\lvert\alpha\rvert=-\alpha$ because they are vertical angles. Therefore, the gradient is $\displaystyle \tan\alpha=\tan(-\angle ACB)=-\tan\angle ACB$. - Feb 26th 2011, 06:22 AMHallsofIvy
Another way to do this, without (directly) calculating the gradient, is to note that, since the distance from the point of tangency to (4, 0) is $\displaystyle \sqrt{21}$, we have, for that angle, C, sin(C)= 2/5 and $\displaystyle cos(C)= \sqrt{21}{5}$. It follows from that that cos(A)= 2/5 and $\displaystyle sin(A)= \sqrt{21}/5$.

From those, we can calculate that the coordinates of point B are $\displaystyle (2 (2/5), 2(\sqr{21}/5)$ and then we can find the equation of the line using the "two point" formula. - Feb 26th 2011, 01:43 PMbjhopper
Hello emarkarov,

In geometry the slope of a line = rise /run so strictly speaking the tangent at C is not the slope but is the same quantity as can be seen when you erect the altitude on hypot of ABC. I admit that geometry today would allow your solution but would be interested in a few replies.

bjh