Question

Write the length of chord cut-off by 2x - y - 1 = 0 from circle x^2 + y^2 = 2 .​

Answer 1

Answer:

Let cord cut off by y=2x+1 is AB = a units and center of circle is O .perpendicular

drawn on cord AB from O is OD.

Circle is x^2+y^2=2

or (x-0)^2+(y-0)^2= (√2)^2 , O (0,0) and r = (OA)=√(2) units

Now OD = (0–2×0–1)/√(1 +4) = 1/√(5) units.

D is mid point of.AB , DA=DB=AB/2 = a/2 units.

In right angled triangle ODB

DB^2+OD^2=OA^2

(a/2)^2 +[1/(√5)]^2 =(√2)^2

a^2/4+1/5 =2

5a^2+4=40

5a^2=36

a^2 = 36/5

a=6/√5 or ( 6/5).√5.

AB = (6/5).√5. units Answer.

Answer 2

Answer:

Length of the chord will be $\dfrac{6}{\sqrt{5}}$

Step-by-step explanation:

A chord cuts a circle at 2 distinct points.

So, solving these two equations should give us two distinct values of 'x' and 'y'.

$y = 2x - 1$

When the value of 'y' from the equation of line is substituted in the equation of the circle, we get:-

$x^{2}+{(2x-1)}^{2}=2 \\5x^{2}+4x-1=0 \\(x+1)(5x-1)=0 \\x = -1, \ \dfrac{1}{5}$

When, $x=-1, \ y=-1$ and when $x=\dfrac{1}{5}, \ y=\dfrac{7}{5}$

We have co-ordinates for both the points of intersection of the chord on the circle.

We can use the distance formula to find the length of the chord:-

$Length =\sqrt{{\left( \dfrac{1}{5}+1 \right)}^{2} +{\left( \dfrac{7}{5}+1 \right)}^{2}} \\Length = \sqrt{\dfrac{180}{25}}=\dfrac{6}{\sqrt5}$