Question

Find the co-ordinates of the middle point of the chord which the circle $x^2+y^2-2x-2=0$ cuts off on the line $y=x-1$

Answer 1

We have,

Equation of line $y=x-1$ …… (1)

Equation of circle $x^2+y^2-2x-2=0$ …… (2)

Substitute the value of $y$ by equation (1) in equation (2) and we get,

$x^2+y^2-2x-2=0$

$x^2+{(x-1)}^2-2x-2=0$

$x^2+x^2+1-2x-2=0$

$2x^2-2x-1=0$

Comparing that $a{x}^2+bx+c=0$ and we get,

$a=2,b=-2,c=-1$

Then we know that,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-(-2)\pm \sqrt{{(-2)}^2-4\times 2\times (-1)}}{2\times 2}$

$=\frac{2\pm \sqrt{12}}{4}$

$=\frac{2\pm 2\sqrt{3}}{4}$

$=\frac{1\pm \sqrt{3}}{2}$

Then, $x_1=\frac{1+\sqrt{3}}{2}$ and $x_2=\frac{1-\sqrt{3}}{2}$

Put the value of $x$ in equation (1) and we get,

$y=\frac{1\pm \sqrt{3}}{2}-1$

$y=\frac{1\pm \sqrt{3}-2}{2}$

$y=\frac{-1\pm \sqrt{3}}{2}$

Then, $y_1=\frac{-1+\sqrt{3}}{2}$ and $y_2=\frac{-1-\sqrt{3}}{2}$

Let the middle point of chord.

$(x^{\prime },y^{\prime })=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$(x^{\prime },y^{\prime })=(\frac{\frac{1+\sqrt{3}}{2}+\frac{1-\sqrt{3}}{2}}{2},\frac{\frac{-1+\sqrt{3}}{2}+\frac{-1-\sqrt{3}}{2}}{2})$

$(x^{\prime },y^{\prime })=(\frac{\frac{1+\sqrt{3}+1-\sqrt{3}}{2}}{2},\frac{\frac{-1+\sqrt{3}-1-\sqrt{3}}{2}}{2})$

$(x^{\prime },y^{\prime })=(\frac{\frac{2}{2}}{2},\frac{\frac{-2}{2}}{2})$

$(x^{\prime },y^{\prime })=(\frac{1}{2},-\frac{1}{2})$

Hence, this is the answer.