# Find the length of the chord joining the points in which the straight line $\frac{x}{a} + \frac{y}{b}= 1$ meets the circle $x^2+y^2=r^2$

Question: Find the length of the chord joining the points in which the straight line $\frac{x}{a} + \frac{y}{b}= 1$ meets the circle $x^2+y^2=r^2$

My initial thoughts were rearranging the straight line equation into the form of $y=mx+c$ which means our equation becomes $y= b - \frac{bx}{a}$

So subbing this into our circle equation we get

$$x^2 + (b - \frac{bx}{a})^2 = r^2$$

$$\Leftrightarrow x^2 + b^2 - \frac{2b^2x}{a} + \frac{b^2x^2}{a^2} = r^2$$

$$\Leftrightarrow a^2x^2+a^2b^2- 2b^2ax+b^2x^2-r^2a^2=0$$

$$\Leftrightarrow (a^2+b^2)x^2-(2b^2a)x+a^2(b^2-r^2)=0$$

Now I was thinking of finding $(x_1-x_2)^2$ and $(y_1-y_2)^2$ and finding the length of the chord by using the distance formula but I want to know if I am on the right track.

--- Attempt at it

So now to find the length of the chord we need to find $\sqrt{ (x_1-x_2)^2 + (y_1+y_2)^2}$

Hence by vieta's formulas we get:

$$\Rightarrow x_1+x_2 = \frac{2ab^2}{a^2+b^2} ~~~~{and} ~~~~ \Rightarrow x_1x_2 = \frac{a^2(b^2-r^2)}{a^2+b^2}$$

As $(x_1-x_2)^2 = (x_1+x_2)^2 -4x_1x_2$

$$(x_1-x_2)^2 = \left(\frac{2ab^2}{a^2+b^2}\right)^2 - \frac{4a^2(b^2-r^2)}{a^2+b^2}$$

$$\Leftrightarrow (x_1-x_2)^2 = \frac{4a^2b^4}{(a^2+b^2)^2} - \frac{4a^2(b^2-r^2)}{a^2+b^2}$$

$$\Leftrightarrow (x_1-x_2)^2 = \frac{4a^2}{a^2+b^2} \left( \frac{b^4}{a^2+b^2} + r^2-b^2 \right)$$

$$\Leftrightarrow (x_1-x_2)^2 = \frac{4a^2}{(a^2+b^2)^2} \left( {b^4} + r^2-b^2(a^2+b^2) \right)$$

$$\Leftrightarrow (x_1-x_2)^2 = \frac{4a^2}{(a^2+b^2)^2} \left( {b^4} + (r^2-b^2)(a^2+b^2) \right)$$

$$\therefore (x_1-x_2)^2 = \frac{4a^2}{(a^2+b^2)^2} \left( r^2a^2+r^2b^2-a^2b^2\right)$$

Now to find the corresponding $(y_1-y_2)^2$

As $y= b - \frac{bx}{a} = b(1-\frac{x}{a}) = \frac{b}{a}(a-x)$ then

$$y_1+y_2 = \frac{b}{a}(a-x_1) + \frac{b}{a}(a-x_2)$$

$$\Leftrightarrow y_1+y_2 = \frac{b}{a} (2a - (x_1+x_2))$$

$$\Leftrightarrow y_1+y_2 = \frac{b}{a} (2a - (\frac{2ab^2}{a^2+b^2})$$

$$\Leftrightarrow y_1+y_2 = 2b(1- \frac{b^2}{a^2+b^2}$$

$$\Rightarrow y_1+y_2 = 2b(\frac{a^2}{a^2+b^2})$$

Now to find $y_1y_2$

$$y_1y_2 = \frac{b}{a}(a-x_1)\frac{b}{a}(a-x_2)$$

$$\Leftrightarrow y_1y_2 = \frac{b^2}{a^2}(a^2-x_1a-x_2a+x_1x_2)$$

$$\Leftrightarrow y_1y_2 = \frac{b^2}{a^2}(a^2-a(x_1+x_2) + (x_1x_2))$$

$$\Leftrightarrow y_1y_2 = \frac{b^2}{a^2}(a^2-a(\frac{2ab^2}{a^2+b^2}) + \frac{a^2(b^2-r^2)}{a^2+b^2})$$

$$\Leftrightarrow y_1y_2 = \frac{b^2}{a^2}(a^2-\frac{2a^2b^2}{a^2+b^2}+ \frac{a^2(b^2-r^2)}{a^2+b^2})$$

$$\Leftrightarrow y_1y_2 = b^2(1- \frac{2b^2}{a^2+b^2} + \frac{b^2-r^2}{a^2+b^2})$$

$$\Rightarrow y_1y_2 = \frac{b^2}{a^2+b^2}(a^2-r^2)$$

So $$(y_1-y_2)^2 = (y_1+y_2)^2-4y_1y_2$$

$$\Leftrightarrow (y_1-y_2)^2 = \left( 2b(\frac{a^2}{a^2+b^2}) \right)^2 - \frac{4b^2}{a^2+b^2}(a^2-r^2)$$

$$\Leftrightarrow (y_1-y_2)^2 = 4b^2(\frac{a^4}{(a^2+b^2)^2})- \frac{4b^2}{a^2+b^2}(a^2-r^2)$$

$$\Leftrightarrow (y_1-y_2)^2 = \frac{4b^2}{a^2+b^2}(\frac{a^4}{a^2+b^2} + r^2-a^2)$$

$$\therefore (y_1-y_2)^2 = \frac{4b^2}{(a^2+b^2)^2}(r^2a^2+r^2b^2-a^2b^2)$$

So Length of Chord is

$$L = \sqrt{ (x_1-x_2)^2 + (y_1+y_2)^2}$$

$$L = \sqrt{ \frac{4a^2}{(a^2+b^2)^2} \left( r^2a^2+r^2b^2-a^2b^2\right) + \frac{4b^2}{(a^2+b^2)^2}(r^2a^2+r^2b^2-a^2b^2)}$$

$$\Leftrightarrow L = \sqrt{ \left(\frac{4}{(a^2+b^2)^2}\right)(r^2a^2+a^2b^2r^2-a^4b^2+r^2a^2b^2+b^2b^4-a^2b^4) }$$

$$\Leftrightarrow L = \frac{2}{a^2+b^2} \sqrt{2a^2b^2r^2 + r^2a^2-a^4b^2+b^2b^4-a^2b^4 }$$

• Are there any restrictions on $a$ and $b$ to ensure you get an intersection? – Ian Miller Jun 18 '16 at 11:35
• Draw it. It then becomes easy – samerivertwice Jun 18 '16 at 11:45
• But yes, your method will work. Solve the quadratic for $x$ first and then substitute into your straight line to find the unique $y$ for each $x$, then find the distance. – samerivertwice Jun 18 '16 at 11:54
• I've added a new, simpler solution to my answer. – PM 2Ring Jun 19 '16 at 5:15

Using Vieta's formula should help.

The distance between the line and the origin is $$\frac{|-ab|}{\sqrt{a^2+b^2}}$$ so, in the following, we may suppose that $$\frac{|-ab|}{\sqrt{a^2+b^2}}\le r.$$ Let $\alpha,\beta$ be the $x$ coordinates of the intersection points.

Then, by Vieta's formula, $$\alpha+\beta=\frac{2ab^2}{a^2+b^2},\quad \alpha\beta=\frac{a^2(b^2-r^2)}{a^2+b^2}$$

Using this, the length is given by \begin{align}\sqrt{(\alpha-\beta)^2+\left(\left(b-\frac{b\alpha}{a}\right)-\left(b-\frac{b\beta}{a}\right)\right)^2}&=\sqrt{\left(1+\left(-\frac ba\right)^2\right)(\alpha-\beta)^2}\\&=\sqrt{\frac{a^2+b^2}{a^2}((\alpha+\beta)^2-4\alpha\beta)}\end{align}

Therefore, the length is

$$2\sqrt{r^2-\frac{a^2b^2}{a^2+b^2}}$$

• Hey thanks for showing me this it really helped, I just wrote up my attempt could you tell me where I went wrong and as to why I did not get the same Length as you.. – bigfocalchord Jun 18 '16 at 12:21
• @dydxx: You are almost there. You have an error at the last step. You should have $$L=\sqrt{\left(\frac{4}{a^2+b^2}\right)^2(r^2a^{\color{red}{4}}+a^2b^2r^2-a^4{b^2}+r^2a^2b^2+\color{red}{r^2}b^4-a^2b^4)}$$ – mathlove Jun 18 '16 at 12:33
• @dydxx In your working on your second last line you should get a $a^4r^2$ and $b^4r^2$ terms when you multiple in the $a^2$ and $b^2$ parts. – Ian Miller Jun 18 '16 at 12:36
• @dydxx: Then, we can write it as $$L=2\sqrt{\frac{(a^4+2a^2b^2+b^4)r^2-a^2b^2(a^2+b^2)}{(a^2+b^2)^2}}.$$ – mathlove Jun 18 '16 at 12:54
• @dydxx: Sorry, I just found an error in my first comment. I wanted to write $$L=\sqrt{\frac{4}{(a^2+b^2)^2}(r^2a^4+a^2b^2r^2-a^4b^2+r^2a^2b^2+r^2b^4-a^2b^4)}$$ – mathlove Jun 18 '16 at 12:57

As Robert Frost's hint suggests, there's a simple geometric solution.

$$\frac{x}{a} + \frac{y}{b}= 1$$

is the intercept form of the equation of a line, i.e., the X & Y intercepts of the line are $a$ and $b$, respectively. Let $c$ be the length of the line segment between the intercept points. This segment is the hypotenuse of a right triangle, so

$$c^2 = a^2 + b^2$$

Let $h$ be the altitude of that triangle, i.e., the perpendicular distance from the origin $O$ to the line. Clearly,

$$hc = ab$$

because $hc$ and $ab$ are both twice the area of that right triangle.

so $$h^2 = \left(\frac{ab}{c}\right)^2 = \frac{a^2b^2}{a^2+b^2}$$

Let $P_0$ and $P_1$ be the points where the line cuts the circle $x^2 + y^2 = r^2$. The triangle $OP_0P_1$ is isosceles since the lengths $OP_0 = OP_1 = r$, and its altitude is $h$. The base of this isosceles triangle is the desired chord. Let $d$ be the length of the chord. By Pythagoras' theorem:

\begin{align} \left(\frac{d}{2}\right)^2 & = r^2 - h^2\\ & = r^2 - \frac{a^2b^2}{a^2+b^2} \end{align}

Thus $$d = 2\sqrt{r^2 - \frac{a^2b^2}{a^2+b^2}}$$

I have created an SVG diagram for this problem but unfortunately this site doesn't support SVG (AFAICT), so here's a PNG version.

$Y$ and $X$ are the intercepts, $P_0P_1$ is the desired chord, $OQ$ is perpendicular to $P_0P_1$.

Solving these two simultaneously gives me:

$$x=\frac{a(b^2\pm \sqrt{r^2(a^2+b^2)-b^2})}{a^2+b^2}$$

$$y=\frac{b(a^2\pm \sqrt{r^2(a^2+b^2)-a^2})}{a^2+b^2}$$

To find the chord length we can find $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

This gives:

$$\sqrt{\left(\frac{2a\sqrt{r^2(a^2+b^2)-b^2}}{a^2+b^2}\right)^2+\left(\frac{2b\sqrt{r^2(a^2+b^2)-a^2}}{a^2+b^2}\right)^2}$$

$$=\sqrt{\frac{4(r^2(a^2+b^2)-a^2b^2)}{a^2+b^2}}$$

$$=2\sqrt{r^2-\frac{a^2b^2}{a^2+b^2}}$$

Also because I didn't read the question properly you could also find the arc length consider the dot product of the two vectors:

$x_1 x_2+y_1 y_2=r^2\cos\theta$

$$\left(\frac{a(b^2+\sqrt{r^2(a^2+b^2)-ab^2})}{a^2+b^2}\right)\left(\frac{a(b^2-\sqrt{r^2(a^2+b^2)-b^2})}{a^2+b^2}\right)+\left(\frac{b(a^2+\sqrt{r^2(a^2+b^2)-a^2})}{a^2+b^2}\right)\left(\frac{b(a^2-\sqrt{r^2(a^2+b^2)-a^2})}{a^2+b^2}\right)=r^2\cos\theta$$

This simplifies easily due to $(x+y)(x-y)=x^2-y^2$

$$\frac{2a^2b^2-r^2(a^2+b^2)}{a^2+b^2}=r^2\cos\theta$$

Hence:

$$\cos\theta=\frac{2a^2b^2-r^2(a^2+b^2)}{(a^2+b^2)r^2}$$

$$\cos\theta=\frac{2a^2b^2}{(a^2+b^2)r^2}-1$$

• The OP wants the chord length, not the arc length. – PM 2Ring Jun 18 '16 at 12:15
• Oh shoot. And I thought it was such an elegant solution. – Ian Miller Jun 18 '16 at 12:26

Here's a slightly different way. Rather than messing around with $x$ and $y$ we express the equation of the line in parametric form and work with the parameter $t$.

The equation of the line is given in intercept form, i.e., the line passes through $(0, b)$ and $(a, 0)$. We can easily change it to parametric form:

$$x = at, \, y = b(1 - t)$$

where $t = 0$ maps to $(0, b)$ and $t = 1$ maps to $(a, 0)$.

The distance between 2 points $t_0$ and $t_1$ is

\begin{align} d & = \sqrt{(x_1-x_0)^2 + (y_1-y_0)^2}\\ & = \sqrt{a^2(t_1-t_0)^2 + b^2((1-t_1) - (1-t_0))^2}\\ & = \sqrt{(a^2 + b^2)(t_1-t_0)^2}\\ d & = \sqrt{a^2 + b^2}|t_1-t_0|\\ \end{align}

Thus we can get the chord length from a nice linear equation of the parameters of the intersection points without having to calculate $x$ and $y$.

To find those parameters we substitute the parametric equations for $x$ and $y$ into the equation of the circle:

$$a^2t^2 + b^2(1-t)^2 = r^2\\ a^2t^2 + b^2 -2b^2t + b^2t^2 = r^2\\ (a^2 + b^2)t^2 -2b^2t + (b^2 - r^2) = 0\\$$

Now the solutions of the general quadratic $At^2 + Bt + C = 0$ are given by $$t = \frac{-B \pm\sqrt{B^2 - 4AC}}{2A}$$ so the absolute difference of the two solutions is $$|t_1-t_0| = \frac{\sqrt{B^2 - 4AC}}{A}$$

Thus \begin{align} d & = \sqrt{a^2 + b^2}\left(\frac{\sqrt{(-2b^2)^2 - 4(a^2 + b^2)(b^2 - r^2)}}{a^2 + b^2}\right)\\ & = \frac{\sqrt{4b^4 - 4(a^2 + b^2)(b^2 - r^2)}}{\sqrt{a^2 + b^2}}\\ & = 2\sqrt{\frac{b^4 - a^2b^2 + a^2r^2 - b^4 + b^2r^2}{a^2 + b^2}}\\ & = 2\sqrt{\frac{a^2r^2 + b^2r^2 - a^2b^2}{a^2 + b^2}}\\ d & = 2\sqrt{r^2 - \frac{a^2b^2}{a^2 + b^2}}\\ \end{align}