circles
Given a circle centered at the origin with radius $5$, find the parametric equation of the tangent line to the circle at the point $(3, −4)$.
The final equation is supposed to be like this: $$(3,-4) + t(x,y).$$
The problem is that I don't know what are the values of $x$ and $y$. Help, please.
The center of the circle can be figured out by the given equation of the circle and is the point : $$C(2,2)$$
But then, if the given line is tangent to your circle, it means that the distance from the center of the circle should be exactly $5$. Manipulating the line equation you have derived, we can yield : $4y + 3x + k = 0$ and then by solving the distance formula, you can yield the exact equation (there will be 2 parallel and diametrically opposite equations thus two tangent lines) :
$$ \left|\frac{Ax_0 + By_0 + Γ}{\sqrt{A^2 +B^2}} \right| = d(P,ε) \Rightarrow \left|\frac{4\cdot 2 + 3 \cdot 2 + k}{\sqrt{4^2+3^2}}\right| = 5$$
$$\Leftrightarrow$$
$$|8 + 6 + k| = 25 \Leftrightarrow \dots$$
Another approach would be substituting the line equation for $x$ and $y$ into your circle's equation and then demanding the equation to have a unique solution, since a tangent line will only have one common point with a circle.
Note : This only works for the case of the circle, when a tangent line can never have $2$ common points. This is not the case for other curves though.
Let us rewrite the issue in the following way :
Being given two lines $(L_1)$ and $(L_2)$, find a general representation $x_m,y_m$ for the center and $R_m$ for the radius of any circle which is tangent to both of them ($m$ being a parameter). This tangency occurs either in the 2 smallest sectors corresponding to the acute angle between $(L_1)$ and $(L_2)$ (as is the case of Fig. 1) or in the 2 largest ones (left empty in this case).
We will consider the case where the intersection point of the two lines is the origin. If this is not the case, one has only to compute the intersection point and make coordinates' translation.
The main idea in this case (lines intersecting at the origin) is that it suffices to know a single circle (called prototype circle, depicted in red on Fig. 1) tangent to both lines and then homothetize it ( = enlarge it).
Fig. 1 : An example with lines whose polar angles are $\pi/12$ and $\pi/4$. The "prototype circle" is the red one.
Here is the Matlab program that has generated Fig. 1 :
clear all;close all;hold on;axis equal
t1=pi/12;a=cos(t1);b=sin(t1);% t1 = polar angle of line 1
t2=pi/4;c=cos(t2);d=sin(t2);% t2 = polar angle of line 2
R=(a-c)/(b+d);% radius (other expression R=(d-b)/(a+c))
x0=a-R*b;y0=b+R*a;% center coordinates of the proto. circle
t=0:0.01:2*pi;
for h=-2:0.1:2; % homothety coefficient
co='b';
if h==1;co='r';end;
plot(h*(x0+R*cos(t)),h*(y0+R*sin(t)),'color',co);
end;
H=2.5;
plot(H*[-a,a],H*[-b,b],'b'); % first line
plot(H*[-c,c],H*[-d,d],'b'); % second line
plot([-3,3],[0,0]); % x-axis
plot([0,0],[-2,2]); % y-axis
Explanation : the important thing to understand is what are lines 4 and 5 made for.
Indeed these lines express the fact that the coordinates of the center of the prototype circle is
$$\binom{x_0}{y_0}=\underbrace{\binom{a}{b}}_{V}+R\underbrace{\binom{-b}{\ \ a}}_{V^{\perp}}\tag{1}$$
with
$$R=\frac{a-c}{b+d}=\frac{d-b}{a+c}\tag{2}$$
(the second equality is readily checked as a consequence of the fact that $a^2+b^2=c^2+d^2=1$).
The proof of the first equality in (2) can be given using trigonometric formulas :
$$\frac{a-c}{b+d}=\frac{\cos(\theta_1)-\cos(\theta_2)}{\sin(\theta_1)+\sin(\theta_2)}=\frac{-2\sin(\tfrac12(\theta_1+\theta_2))\sin(\tfrac12(\theta_1-\theta_2))}{2\sin(\tfrac12(\theta_1+\theta_2))\cos(\tfrac12(\theta_1-\theta_2))}$$
Thus
$$R=\tan(\tfrac12\theta_3) \ \text{with} \ \theta_3:=\theta_2-\theta_1,\tag{3}$$
(3) being true (see Fig. 2 for understanding it).
Fig. 2
Best Answer
Guide:
The radius is perpedicular to the tangent line.
Can you find a solution to $$3x-4y=0?$$