# 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.

• Please show the own effort to solve the problem, it is the way it works, potential answerers can then economically give pointed answers. Sep 9, 2019 at 22:18
• You need to solve for $x$. Use the quadratic formula. It'd probably be easier if you didn't expand. Sep 11, 2019 at 0:24

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.

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 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.

$$\large \frac{y_1^2 x^2 - y_1 x_1 x + x_1^2 x^2 - x_1^2 x + 0.5 x_1^2}{x_1^2} = \normalsize 0.25$$

$$(x_1^2+y_1^2)x^2 - x_1(x_1+y_1)\;x + 0.25\;x_1^2 = 0$$

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.

$$(x_1-{1\over2})^2 + (y_1-{1\over2})^2 = {1\over4}$$

$$x_1^2 + y_1^2 = x_1 + y_1 - {1 \over 4}$$

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