Question

The length of the chord joining the points in which the straight line $\frac{x}{3}+\frac{y}{4}=1$ circle $x^2+y^2$ = $\frac{169}{25}$

• A. 1
• B. 2
• C. 4
• D. 8

Answer 1

Answer: (B)

We have, equations of line and circle are
$\frac{x}{3}+\frac{y}{4}=\mathrm{1....}$(i)
and $\Rightarrow 25x^2+25y^2=\mathrm{169....}$(ii)

$\therefore$ Length of the perpendicular from centre $(0,0)$ of circle to the line (i) $=12/5$ and radius of the circle $=13/5$
$\therefore$ Required length $=2\sqrt{\frac{169}{25}-\frac{144}{25}}=2$