Question

A circle given by x^2+y^2=25 is cut into two segments by the line 2x-3y=-1.Calculate the lenght of the cord when the line cuts the circle.?

Answer 1

Length of chord is `9.985`.

Explanation:

Find the points of intersection by solving equations simultaneously and then distance betwen them.

The equations are `x^2+y^2=25` and `2x-3y=-1`

From secondequation `x=(3y-1)/2` and putting this in first equation, we get

`((3y-1)/2)^2+y^2=25`

or `9y^2-6y+1+4y^2=100`

or `13y^2-6y-99=0`

and using quadratic formula

`y=(6+-sqrt(36-4*13*(-99)))/26`

= `(6+-sqrt5184)/26=(6+-72)/26`

i.e. `y=3` or `-33/13`

and corresponding `x=4` or `-99/26-1/2=(-99-13)/26=-56/13`

Hence points of intersection are `(4,3)` and `(-56/13,-33/13)`

and length of chord is `sqrt((4+56/13)^2+(3+33/13)^2)`

= `1/33sqrt(108^2+72^2)`

= `36/13sqrt13=9.985`

graph{(x^2+y^2-25)(2x-3y+1)=0 [-10.125, 9.875, -4.88, 5.12]}

Answer 2

The given equation of the circle is `x^2+y^2=5^2`.

So the coordinates of its center `(O)` is `(0,0)` and length of its radius is `5` unit.

The equation of the given chord `AB` is `2x-3y+1=0`,where A and B represent the points of intersection of the chord with the circle.

Hence length of the perpendicular `(OC)` drawn from `O` to AB is given by

`OC=(2*0-3*0+1)/sqrt(2^2+3^2)=1/sqrt13`

`DeltaOAC` is a right triangle. Here `OA=5`.

So by Pythagoras theorem

`AC=sqrt(OA^2-OC^2)`

`=>AC=sqrt(5^2-1/13)=18/sqrt13`

So length of the chord

`AB=2AC=2xx18/sqrt13=36/13 sqrt13`