Question

The length of the chord intercepted by the $x^2+y^2=r^2$ on the line $\frac{x}{a}+\frac{y}{b}=1$ is-

• A. $\frac{2\sqrt{r^2(a^2+b^2)-a^2{b}^2}}{a^2+b^2}$
• B. $2\sqrt{\frac{r^2(a^2+b^2)-a^2{b}^2}{a^2+b^2}}$
• C. $\sqrt{\frac{r^2(a^2+b^2)-a^2{b}^2}{a^2+b^2}}$
• D. None of these

Answer 1

The correct option is A $\frac{2\sqrt{r^2(a^2+b^2)-a^2{b}^2}}{a^2+b^2}$

$\frac{x}{a}+\frac{y}{b}=1$ and $x^2+y^2=r^2$ $y=-\frac{bx}{a}+b$ substitute this in equation of circle we get $x^2+{(-\frac{bx}{a}+b)}^2=r^2$ $x^2(1+\frac{b^2}{a^2})-\frac{2b^{2}x}{a}+b^2-r^2=0$ $\Rightarrow$ $x=\frac{\frac{2b^2}{a}\pm \sqrt{\frac{4b^4}{a^2}-4b^2-\frac{4b^4}{a^2}+4r^2+\frac{4r^2{b}^2}{a^2}}}{2(1+\frac{b^2}{a^2})}$ by Sridharacharya formula On simplifying we get $x=\frac{b^{2}a\pm a\sqrt{r^2(a^2+b^2)-a^2{b}^2}}{a^2+b^2}$ Now $\frac{x}{a}+\frac{y}{b}=1$ Hence $y=\frac{a^{2}b\pm \sqrt{r^2(a^2+b^2)-a^2{b}^2}}{-(a^2+b^2)}$ Now $length=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$ On applying this formula we get $l^2=\frac{4(a^2+b^2)\left[r^2\right(a^2+b^2)-a^2{b}^2]}{(a^2+b^2{)}^2}$ Hence $l=2\sqrt{\frac{\left[r^2\right(a^2+b^2)-a^2{b}^2]}{(a^2+b^2)}}$