# Thread: Circle Geometry

1. ## Circle Geometry

The point A has coordinates $(2, -1)$. Find the equation of the chord of the circle which has $A$ as its midpoint.

I don't know how to do it with limited information. Could somebody help please? I'm a bit shoddy at these questions.

It has preceding questions which say let $C$ be the circle with equation $x^2 + y^2 - 2x + 4y - 11 = 0$

I found the centre and radius.

Then it asked me to determine whether the origin lies inside or outside C; it lies inside.

2. As written there are infinitely many answers.
Therefore, you have posted only a very small part of the complete question.
Please revise the question giving all the information.

3. Well the centre of $C$ is $(1, -2)$ and the radius is 4.

I found that the origin lies inside the circle since the distance to the origin from the centre is only $\sqrt 5$ which is smaller than 4.

They are the only preceding pieces of information I have.

4. The slope of the line determined by $A~\&~C$ is $\dfrac{(-2)-(-1)}{(1)-(2)}=1$.
Therefore the slope of the chord is $-1$. WHY?

5. Originally Posted by Plato
The slope of the line determined by $A~\&~C$ is $\dfrac{(-2)-(-1)}{(1)-(2)}=1$.
Therefore the slope of the chord is $-1$. WHY?
Since the chord is perpendicular to the tangent gradient you've found?

But how is AC a tangent gradient?

6. Originally Posted by Femto
Since the chord is perpendicular to the tangent gradient you've found?
But how is AC a tangent gradient?
The line determined by the center of a circle and the midpoint of a chord is the perpendicular bisector of the chord.

7. Oh right thanks very much I forgot about that!