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 $x^2+y^2=r^2$.

Answer 1

Equation of circle is $x^2+y^2=r^2$

Let the line $\frac{x}{a}+\frac{y}{b}=1$ intersect the circle at $A$ and $B$

Draw $OC$ perpendicular to $AB$

$OC=\frac{|0+0-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=\frac{ab}{\sqrt{a^2+b^2}}$

In $△OAC$ $,$ ${OA}^2={AC}^2+{OC}^2$

$\Rightarrow r^2={\left(\frac{ab}{\sqrt{a^2+b^2}}\right)}^2+{AC}^2$

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

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

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

length of chord $=AB=2AC$

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