Question

Find the length of the chord intercepted by the circle $x^2+y^2-8x-2y-8=0$ on the line $x+y+1=0$.

Answer 1

The equation of the circle can be written as

$(x-4{)}^2+(y-1{)}^2-16-1-8=0$

$(x-4{)}^2+(y-1{)}^2=25$.

Now the points of intersection of the line and the circle can be found out by substituting the equation of the line in the circle.

$y=-(x+1)$

Hence

$(x-4{)}^2+(-x-1-1{)}^2=25$

$(x-4{)}^2+(-x-2{)}^2=25$

$(x^2-8x+16)+(x^2+4x+4)=25$

$2x^2-4x+20=25$

$2x^2-4x-5=0$

Therefore

$x_1=\frac{2+\sqrt{14}}{2}$ and $x_1=\frac{2-\sqrt{14}}{2}$

Similarly $y_1=\frac{-4-\sqrt{14}}{2}$ and $y_2=\frac{-4+\sqrt{14}}{2}$

Hence the length of the chord is

$D=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$

$=\sqrt{(\sqrt{14}{)}^2+(\sqrt{14}{)}^2}$

$=\sqrt{2\times 14}$

$=\sqrt{28}$

$=2\sqrt{7}$ units.