A line goes through a point outside a circle and a point on the circle. Find the second intersection point between the line and the circle Say, I have a circle 
$$\left(x – \frac 12 \right)^2 + \left( y  – \frac 12 \right)^2 = \frac 14$$
If I have a point $(0,0)$ and a point on the circumference of the circle $(x_1,y_1)$, how would I find the second intersection point between the circle and the line that goes through these two points $(0,0)$ and $(x_1,y_1)$? 
I wrote the equation of the line 
$$y=\frac{y_1}{x_1}x$$
and tried substituting it into the equation of the circle. After expanding everything I got,
$$\frac{y_1^2 x^2 - y_1 x_1 x + x_1^2 x^2 - x_1^2 x + 0.5 x_1^2}{x_1^2} = 0.25$$
But, I'm not sure how to proceed from here.
 A: The cartesian equation of the line passing by $(x_1,y_1)$ and $(x_2,y_2)$ is $y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1$, let $(x,y)\in\mathcal{C}\cap\mathcal{D}$, plot this into the equation of the circle : $(x-h)^2+\left(\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1-k\right)^2=r^2$. It is a second order equation of the variable $x$.
A: You tried to find the second point by solving the system of equations, which could be a bit messy because of the quadratic equation involved.
A cleaner approach is to find the mid-point along the chord that connects the two points $(x_1,y_1)$ and $(x_2,y_2)$. The line of the chord is 
$$y = \frac{y_1}{x_1} $$
And the line perpendicular to the chord and going through the center of the circle is
$$y-\frac 12 = -\frac{x_1}{y_1} \left( x - \frac 12  \right)$$
Next, find the intersection point of the two lines, 
$$x_m = \frac 12 \frac{y_1+x_1}{y_1^2+x_1^2}x_1, \>\>\> 
y_m = \frac 12 \frac{y_1+x_1}{y_1^2+x_1^2}y_1,$$
Since $(x_m,y_m)$ is the midpoint of $(x_1,y_1)$ and $(x_2,y_2)$, simply take the average below,
$$x_m = \frac {x_1+x_2}{2}, \>\>\> y_m = \frac {y_1+y_2}{2}$$
The second point $(x_2,y_2)$ can then be obtained, 
$$x_2= 2x_m - x_1 = \left( \frac{y_1+x_1}{y_1^2+x_1^2}-1 \right)x_1 $$
$$y_2= 2y_m - x_1 = \left( \frac{y_1+x_1}{y_1^2+x_1^2}-1 \right)y_1 $$
Recognizing that $(x_1,y_1)$ satisfies 
$$x_1^2+y_1^2-x_1-y_1+\frac 14=0$$
The above expressions for $(x_2,y_2)$ could be further simplified.
A: A point $P'$ on the line through $O=(0,0)$ and $P=(p,q)$ is given by $P' = (tp, tq)$ for some (possibly negative) $t$.
Now, the power of $O$ with respect to the circle with center $K=(h,k)$ and radius $r$ is defined as
$$\operatorname{pow}_KO := |OK|^2 - r^2 \;= h^2+k^2-r^2 \tag{1}$$
By the Power of a Point Theorem, we also have
$$\operatorname{pow}_KO = |OP||OP'| \;= t\,|OP|^2 = t\left(p^2+q^2\right) \tag{2}$$
Therefore,

$$t = \frac{h^2+k^2-r^2}{p^2+q^2} \tag{$\star$}$$

The reader can show that, in the case where $h=k=r=1/2$, this answer is consistent with @Quanto's solution.
A: Using your last expression, rewrite it as quadratic:  
$\large \frac{y_1^2 x^2 - y_1 x_1 x + x_1^2 x^2 - x_1^2 x + 0.5 x_1^2}{x_1^2} = \normalsize 0.25$
$(x_1^2+y_1^2)x^2 - x_1(x_1+y_1)\;x + 0.25\;x_1^2 = 0$
Product of roots, $\large x_1x_2 = {0.25 x_1^2 \over x_1^2+y_1^2} →  x_2 = {x_1 \over 4 (x_1^2+y_1^2)}$
We could simplify further, since the point is on the circle.
$(x_1-{1\over2})^2 + (y_1-{1\over2})^2 = {1\over4}$ 
$x_1^2 + y_1^2 = x_1 + y_1 - {1 \over 4} $
$$x_2 = {x_1 \over 4(x_1+y_1)-1}$$
$$y_2 = {y_1 \over 4(x_1+y_1)-1}$$
