Question

Find the point of a intersection of circle $x^2+y^2=25$ and line $4x+3y=12$ and also find length of intersecting chord.

Answer 1

$x^2+y^2=25$

$4x+3y=12$
$x=\frac{12-3y}{4}=3-\frac{3y}{4}$

${(3-\frac{3y}{4})}^2+y^2=25$

$9+\frac{9y^2}{16}+y^2-\frac{9y}{2}=25$

$25y^2+144-72y=400$
$25y^2-72y-256=0$

$y=4.949,-2.069$
$x=-0.711,4.551$

length of chord $=2(r^2-(\perp distance{)}^2)$

$\perp distance=\frac{|-12|}{\sqrt{4^2+3^2}}=\frac{12}{5}$

Length of chord =$2(5^2-{\left(\frac{12}{5}\right)}^2)$

$2(25-\frac{144}{25})$

$=\frac{481\times 2}{25}=\frac{962}{25}$

$=38.48$units.