Find chord length given equations for circle and line

Question

Is there any way to find the length of a chord if you are given the equations for the chord and the circle? Say you have a circle with an equation of $x^2 + y^2 = 4$ and the line $y = 3x - 4$. Is there any way to get the chord length from this?

I would assume this may be possible by substituting $3x-4$ for $y$ in the circle equation and solving for $x$ to get the $x$ values of the points of intersection, and then repeating this process but instead substituting $(y+4)/3$ for x, and solving for y to get the y values of these points. Then would it be possible to simply use the distance forumla to get the length of the chord?

Thanks in advance for any help.

Answer 1

The straight line meets the circle in $A$ and $B$. Calculate the intersection of $y=3x-4$ with the perpendicular line through the origin $O$, namely $y=-\frac13x$, it's $C(1.2,-0.4)$. Now consider the right-angled triangle $OAC$ to determine the length of $AC$ as $\sqrt{2.4}$. Hence the chord has length $2\sqrt{2.4}$.

Answer 2

Another way can be found to calculate the intersection points: $$x^2+(3x-4)^2=4$$ For your work: We find $$\left\{\left\{x\to \frac{1}{5} \left(6-\sqrt{6}\right),y\to \frac{1}{5} \left(-2-3 \sqrt{6}\right)\right\},\left\{x\to \frac{1}{5} \left(6+\sqrt{6}\right),y\to \frac{3 \sqrt{6}}{5}-\frac{2}{5}\right\}\right\}$$

Answer 3

Following @amd's hint, the squared distance of the origin to the line is

$$\frac{4^2}{1^2+3^2}$$

and by Pythagoras the chord length is

$$2\sqrt{4-\dfrac{16}{10}}=4\sqrt{\frac 35}.$$

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Alternatively, a parametric equation of the line is

$$\begin{cases}x=t,\\y=3t-4.\end{cases}$$

We normalize the coefficients so that $t$ represents the abscissa along the line

$$\begin{cases}x=\dfrac t{\sqrt{10}},\\y=\dfrac{3t}{\sqrt{10}}-4\end{cases}$$

and solve

$$x^2+y^2=\frac{t^2}{10}+\left(\frac{3t}{\sqrt{10}}-4\right)^2=4$$ which is quadratic in $t$. The chord length is the difference between the roots, $\sqrt{\Delta}$.