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

Question

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.

Answer 1

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.

Answer 2

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 $x$.

Answer 3

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.

Answer 4

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} $

$$x_2 = {x_1 \over 4(x_1+y_1)-1}$$ $$y_2 = {y_1 \over 4(x_1+y_1)-1}$$