# Thread: locus of mid point of line inersecting circle

1. ## locus of mid point of line inersecting circle

A circle passes through the points A, B, and C which have coordinates $(0,3), (\sqrt{3},0), (-\sqrt{3},0)$ respectively.
Equation of circle $x^2+y^2-2y=3$
A line y=mx-3of variable gradient m, cuts the circle at L and M. Find the Cartesian equation of the locus of the midpoint of the line LM.
I found the values of x and y in terms of m.
$(\frac{4m+2\sqrt{m^2-3}}{m^2+1}, \frac{m^2+2m\sqrt{m^2-3}-3}{m^2+1})$ $(\frac{4m-2\sqrt{m^2-3}}{m^2+1}, \frac{m^2-2m\sqrt{m^2-3}-3}{m^2+1})$
Now what's next? I don't know hot to proceed from here. Any help is greatly appreciated! Thanks!

2. "A circle passes through the points A, B, and C which have coordinates respectively. Equation of circle
A line y=mx-3of variable gradient m, cuts the circle at L and M. Find the Cartesian equation of the locus of the midpoint of the line LM."

How i do it . . . .

x^2 + y^2 - 2y = 3 - - - - (1)

and y = mx + 3 - - - - (2) , substituting
_________________________________________

x^2 + (mx + 3)^2 - 2(mx + 3) = 3,

x^2 + m^2x^2 + 6mx + 9 - 2mx - 6 = 3,

(1 + m^2)x^2 + 4mx = 3 - 9 + 6 = 0,

(1 + m^2)x^2 + 4mx = 0

x[(1 + m^2)x + 4m] = 0,

x = 0, then y = 3

x = -4m/(1 + m^2), then y = (3 - m^2)/(1 + m^2)

i don't know what to continue either . . ..

3. Hello arze
Originally Posted by arze
A circle passes through the points A, B, and C which have coordinates $(0,3), (\sqrt{3},0), (-\sqrt{3},0)$ respectively.
Equation of circle $x^2+y^2-2y=3$
A line y=mx-3of variable gradient m, cuts the circle at L and M. Find the Cartesian equation of the locus of the midpoint of the line LM.
I found the values of x and y in terms of m.
$(\frac{4m+2\sqrt{m^2-3}}{m^2+1}, \frac{m^2+2m\sqrt{m^2-3}-3}{m^2+1})$ $(\frac{4m-2\sqrt{m^2-3}}{m^2+1}, \frac{m^2-2m\sqrt{m^2-3}-3}{m^2+1})$
Now what's next? I don't know hot to proceed from here. Any help is greatly appreciated! Thanks!
The straight line $y=mx-3$ meets $x^2+y^2-2y=3$ where

$x^2 +(mx-3)^2 -2(mx-3)=3$

$\Rightarrow x^2(1+m^2)-8mx+12=0$

If the roots of this equation are $x_1,x_2$, then $x_1+x_2 = \frac{8m}{1+m^2}$

and if the line meets the circle where $y = y_1,y_2$, then $y_1+y_2 = mx_1-3+mx_2 - 3 = m(x_1+x_2)-6=\frac{8m^2}{1+m^2}-6=\frac{2m^2-6}{1+m^2}$

Now the mid-point of the line joining $(x_1,y_1), (x_2,y_2)$ is $\Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big)$. So its locus is given by:

$x=\frac{4m}{1+m^2}$ (1)

$y = \frac{m^2-3}{1+m^2}$ (2)

From (2): $m^2 = \frac{3+y}{1-y}$ (3)

and from (1) and (2): $1+m^2 = \frac{4m}{x}=\frac{m^2-3}{y}$

Subst from (3):

$\frac{4}{x}\sqrt{\frac{3+y}{1-y}}= \frac{\dfrac{3+y}{1-y}-3}{y}=\frac{4y}{1-y}$

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

$\Rightarrow x^2y^2 = (3+y)(1-y)$

Hello arzeThe straight line $y=mx-3$ meets $x^2+y^2-2y=3$ where

$x^2 +(mx-3)^2 -2(mx-3)=3$

$\Rightarrow x^2(1+m^2)-8mx+12=0$

If the roots of this equation are $x_1,x_2$, then $x_1+x_2 = \frac{8m}{1+m^2}$

and if the line meets the circle where $y = y_1,y_2$, then $y_1+y_2 = mx_1-3+mx_2 - 3 = m(x_1+x_2)-6=\frac{8m^2}{1+m^2}-6=\frac{2m^2-6}{1+m^2}$

Now the mid-point of the line joining $(x_1,y_1), (x_2,y_2)$ is $\Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big)$. So its locus is given by:

$x=\frac{4m}{1+m^2}$ (1)

$y = \frac{m^2-3}{1+m^2}$ (2)

From (2): $m^2 = \frac{3+y}{1-y}$ (3)

and from (1) and (2): $1+m^2 = \frac{4m}{x}=\frac{m^2-3}{y}$

Subst from (3):

$\frac{4}{x}\sqrt{\frac{3+y}{1-y}}= \frac{\dfrac{3+y}{1-y}-3}{y}=\frac{4y}{1-y}$

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

$\Rightarrow x^2y^2 = (3+y)(1-y)$

Thanks!
I can't get the answer though, its supposed to be $(x^2+y^2+2y-3)(y+3)=0$

5. Hello arze
Originally Posted by arze
Thanks!
I can't get the answer though, its supposed to be $(x^2+y^2+2y-3)(y+3)=0$
Sorry, I lost a term in $y$ just here:

$\frac{4}{x}\sqrt{\frac{3+y}{1-y}}= \frac{\dfrac{3+y}{1-y}-3}{y}=\frac{4y}{1-y}$
It should be:

$\frac{4}{x}\sqrt{\frac{3+y}{1-y}}= \frac{\dfrac{3+y}{1-y}-3}{y}=\frac{4y}{\color{red}y\color{black}(1-y)}=\frac{4}{1-y}$

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

$\Rightarrow x^2=(3+y)(1-y)$

which is, I think, the correct answer.

I cannot understand why they have written the answer in the form that you quoted:

$(x^2+y^2+2y-3)(y+3)=0$

This is the product of two factors which equals zero; i.e.

$x^2+y^2+2y-3=0$

or

$y+3=0$

The first of these, when re-arranged, is the same as my revised answer. The second is a horizontal line, which, as far as I can see, does not form part of the locus of the mid-point of the chord.

Hello arzeSorry, I lost a term in $y$ just here:

It should be:

$\frac{4}{x}\sqrt{\frac{3+y}{1-y}}= \frac{\dfrac{3+y}{1-y}-3}{y}=\frac{4y}{\color{red}y\color{black}(1-y)}=\frac{4}{1-y}$

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

$\Rightarrow x^2=(3+y)(1-y)$

which is, I think, the correct answer.

I cannot understand why they have written the answer in the form that you quoted:

$(x^2+y^2+2y-3)(y+3)=0$

This is the product of two factors which equals zero; i.e.

$x^2+y^2+2y-3=0$

or

$y+3=0$

The first of these, when re-arranged, is the same as my revised answer. The second is a horizontal line, which, as far as I can see, does not form part of the locus of the mid-point of the chord.