# Find the equation of the straight line

• February 25th 2011, 03:57 AM
Punch
Find the equation of the straight line
Find, without differentiation, the equation of the straight lines which pass through the point $(4,0)$ and are tangential to the circle, $(x+1)^2+y^2=4$
• February 25th 2011, 04:09 AM
emakarov
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.
• February 25th 2011, 04:22 AM
Punch
Sorry what I did was...

Circle center $= (-1,0)$
Let the unknown point of intersection be $(x,y)$

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

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

$\sqrt{(x-4)^2+(y-0)^2}=\sqrt{21}$
$x^2-8x+16+y^2=21$
$x^2+y^2-8x-5=0$
$x^2+y^2=8x+5$

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

$\frac{y-0}{x+1}=\frac{1}{\frac{y-0}{x-4}}$
$\frac{y}{x+1}=\frac{x-4}{y}$
$y^2=(x-4)(x+1)$
$y^2=x^2-3x-4$
sub $y^2=x^2-3x-4,$

$x^2+(x^2-3x-4)=8x+5$
$2x^2-11x+1=0$

With the value of x found, i can find y but the equation of the line i got is different from the answer
• February 25th 2011, 04:41 AM
emakarov
I would do the following. The tangent of the angle at (4,0) is $2/\sqrt{21}$. Therefore, the equations of the two lines are $y=\pm(x-4)\cdot2/\sqrt{21}$.
• February 25th 2011, 05:10 AM
bjhopper
find 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
• February 25th 2011, 11:32 PM
Punch
Quote:

Originally Posted by emakarov
I would do the following. The tangent of the angle at (4,0) is $2/\sqrt{21}$. Therefore, the equations of the two lines are $y=\pm(x-4)\cdot2/\sqrt{21}$.

hi, i didn't understand the part when u said "The tangent of the angle at (4,0) is $2/\sqrt{21}$." I know which lines u are referring to but is there a formula which says $tanA=gradient$?
• February 26th 2011, 01:10 AM
emakarov

Since $\angle ABC=90^\circ$, $\tan\angle BCA=AB/BC=2/\sqrt{21}$. Therefore, the slope of BC is $-2/\sqrt{21}$. Does this answer your question?
• February 26th 2011, 02:57 AM
Punch
Quote:

Originally Posted by emakarov

Since $\angle ABC=90^\circ$, $\tan\angle BCA=AB/BC=2/\sqrt{21}$. Therefore, the slope of BC is $-2/\sqrt{21}$. Does this answer your question?

sorry, but why is $\tan\angle BCA$ the gradient of line BC?
• February 26th 2011, 03:21 AM
bjhopper
Hi,
Emarkarov;s equations are correct but the question is " Is this solution in a geometry problem valid ? "

bjh
• February 26th 2011, 05:32 AM
emakarov
Quote:

Originally Posted by Punch
sorry, but why is $\tan\angle BCA$ the gradient of line BC?

I think, by definition, the gradient of line BC is $\tan\alpha$. (Please correct me if I am wrong.) Taken direction into account, $\alpha<0$. Also, $\angle ACB=\lvert\alpha\rvert=-\alpha$ because they are vertical angles. Therefore, the gradient is $\tan\alpha=\tan(-\angle ACB)=-\tan\angle ACB$.
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 $\sqrt{21}$, we have, for that angle, C, sin(C)= 2/5 and $cos(C)= \sqrt{21}{5}$. It follows from that that cos(A)= 2/5 and $sin(A)= \sqrt{21}/5$.
From those, we can calculate that the coordinates of point B are $(2 (2/5), 2(\sqr{21}/5)$ and then we can find the equation of the line using the "two point" formula.