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