Question

find the length of the chord x-y-3=0 and x2+y2-x+3y-22=0​

Answer 1

Step-by-step explanation:

To find the co-ordinates of the end of the chords, let us substitute the equation of the line in the circle.

Hence

x2+(x−3)2−x+3(x−3)−22=0x2+(x−3)2−x+3(x−3)−22=0

⇒x2+x2−6x+9−x+3x−9−22=0⇒x2+x2−6x+9−x+3x−9−22=0

⇒2x2−4x−22=0⇒2x2−4x−22=0

⇒x2−2x−11=0⇒x2−2x−11=0

⇒x=2±√4+442⇒x=2±4+442

Hence x=1±2√3x=1±23

Therefore x1=1+2√3x1=1+23 and x2=1−2√3x2=1−23.

Hence y1=−2+2√3y1=−2+23 and y2=−2−2√3y2=−2−23

Hence the length of the chord is

D=√(x2−x1)2+(y2−y1)2D=(x2−x1)2+(y2−y1)2

=√(4√3)2+(4√3)2=(43)2+(43)2

=√96=96

=4√6=46 units.