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
$$ L = \sqrt{ (x_1-x_2)^2 + (y_1+y_2)^2}$$
$$L = \sqrt{ \frac{4a^2}{(a^2+b^2)^2} \left( r^2a^2+r^2b^2-a^2b^2\right) + \frac{4b^2}{(a^2+b^2)^2}(r^2a^2+r^2b^2-a^2b^2)}$$
$$ \Leftrightarrow L = \sqrt{ \left(\frac{4}{(a^2+b^2)^2}\right)(r^2a^2+a^2b^2r^2-a^4b^2+r^2a^2b^2+b^2b^4-a^2b^4) } $$
$$ \Leftrightarrow L = \frac{2}{a^2+b^2} \sqrt{2a^2b^2r^2 + r^2a^2-a^4b^2+b^2b^4-a^2b^4 } $$
Best Answer
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