# Math Help - Finding a chord given the midpoint of it's equation

1. ## Finding a chord given the midpoint of it's equation

Given you have the equation of a circle and a point inside it, how do you go about calculating the equation of a chord that has that point as its midpoint?

I have figured that i will need to find the two points on the circumfrence whose x values and y values average to the midpoint. Getting the equation should then be trivial.

To give an example: let circle C be x^2 + y^2 -2x + 4y -11 = 0

Let the desired midpoint,A, be (2,-1)

So x2-x1 = 4 and y2-y1 = -2 (with regard to the points i am looking for on the circumfrence)

and x1 and y1 and x2 and y2 must fit into the equation C.

From here i'm not sure how to solve this problem simultaniously. Is my inital work correct?

2. Hello, StaryNight!

Given you have the equation of a circle and a point inside it, how do you
calculate the equation of a chord that has that point as its midpoint?

To give an example: let circle $C$ be: . $x^2 + y^2 -2x + 4y -11 \:= \:0$

Let the desired midpoint be $A(2,-1)$
Code:
            |
|
| * * *
* | *       *
----*---+---*-------*------
*    |     *   *  *
|       o A(2,-1)
*     |  C  *   *   *
*     |   o       * *
*     | (1,-2)       *
|
*    |            *
*   |           *
* |         *
| * * *
|

We find that the equation of the circle is: . $(x-1)^2 + (y+2)^2 \:=\:16$
. . It has center $C(1,-2)$ and radius 4.

The radius that bisects a chord is perpendicular to the chord.

The slope of $CA$ is: . $m_1 \:=\:\frac{\text{-}1-(\text{-}2)}{2-1} \;=\;1$

. . Hence, the slope of the chord is: . $m_2 \:=\:-1$

Now write the equation of the line through $A(2,-1)$ with slope $m_2 = -1.$

3. Originally Posted by StaryNight
Given you have the equation of a circle and a point inside it, how do you go about calculating the equation of a chord that has that point as its midpoint?

I have figured that i will need to find the two points on the circumfrence whose x values and y values average to the midpoint. Getting the equation should then be trivial.

To give an example: let circle C be x^2 + y^2 -2x + 4y -11 = 0

Let the desired midpoint,A, be (2,-1)

So x2-x1 = 4 and y2-y1 = -2 (with regard to the points i am looking for on the circumfrence)

and x1 and y1 and x2 and y2 must fit into the equation C.

From here i'm not sure how to solve this problem simultaniously. Is my inital work correct?

Your circle is $x^2 + y^2 -2x + 4y -11 = 0$ but this is same as $(x-1)^2+(y+2)^2=16$ So coordinate of the center is (1,-2). We know that "radius"( actually an infinite line that lies on radius) passes through (1,-2) and the midpoint (2,-1) and it is perpendicular to the "chord". Equation of "radius" (knowing two points) is $y=x-3$ now knowing that this line is perpendicular to the "chord", slope of "chord" can be found using famous $-1/m$ where $m=1$ in our case. So equation of the "chord" is $y=-x+b$. We know that "chord" passes through (2,-1) so $b=1$.So finally equation of the "chord" is $y=-x+1$.