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.
Best Answer
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.