# Circle Geometry

• Dec 30th 2010, 07:21 AM
Femto
Circle Geometry
The point A has coordinates $\displaystyle (2, -1)$. Find the equation of the chord of the circle which has $\displaystyle A$ 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 $\displaystyle C$ be the circle with equation $\displaystyle x^2 + y^2 - 2x + 4y - 11 = 0$

I found the centre and radius.

Then it asked me to determine whether the origin lies inside or outside C; it lies inside.
• Dec 30th 2010, 07:29 AM
Plato
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.
• Dec 30th 2010, 07:36 AM
Femto
Well the centre of $\displaystyle C$ is $\displaystyle (1, -2)$ 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 $\displaystyle \sqrt 5$ which is smaller than 4.

They are the only preceding pieces of information I have.
• Dec 30th 2010, 08:26 AM
Plato
The slope of the line determined by $\displaystyle A~\&~C$ is $\displaystyle \dfrac{(-2)-(-1)}{(1)-(2)}=1$.
Therefore the slope of the chord is $\displaystyle -1$. WHY?
• Dec 30th 2010, 08:44 AM
Femto
Quote:

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

Since the chord is perpendicular to the tangent gradient you've found?

But how is AC a tangent gradient?
• Dec 30th 2010, 09:02 AM
Plato
Quote:

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.
• Dec 30th 2010, 09:30 AM
Femto
Oh right thanks very much I forgot about that!