# Geometry of a circle

• Jan 12th 2011, 11:34 AM
Beggarsbelief
Geometry of a circle
Two circles intersect at p(2,0) and q(-2,8). The distance from the centre of each circle to the common chord [pq] is √20

Find the equations of the two circles.

That is the question, I have completed though it took me roughly 40 minutes to complete. I figured out the radius through simple triangle geometry and then created 3 equations in g, f, and c (G being the centre of the circle's x value) (F being the centre of the circle's y value) and c being c.

What resulted was a long drawn out form of substitution which took up one full a4 page to write out and solve, substituting in large equations for f, f^2, c, and c^2 which resulted in getting an equation all in g which eventually led me to a result for g and from there on, substituting back in was easy.

My problem is that it took so long, I am wondering is there another way to solve this question, through trigonometry or did I over complicate it do you think? I am sitting my exams in June and I am expected to answer at least 3 questions of this difficulty along with various other varying difficulty questions, do you think there is an easier way to solve this one than the long, drawn out, painful process I endured, or was what I did correct?
• Jan 12th 2011, 11:47 AM
Also sprach Zarathustra
Yes, I think...

First mark O and T the centers...

Then mark as M the mid of PQ.

Find PM

Find PO and PT

Then use the formula between two points...

hmmm... (sorry for the mess... 1 minute thinking.)
• Jan 12th 2011, 12:02 PM
earboth
Quote:

Originally Posted by Beggarsbelief
Two circles intersect at p(2,0) and q(-2,8). The distance from the centre of each circle to the common chord [pq] is √20

Find the equations of the two circles.

...

Here is what I would have done:

1. Draw a sketch.

Let $M_1(m_{11}, m_{12})$ and $M_2(m_{21}, m_{22})$ denote the midpoints of the circles $C_1$ and $C_2$ respectively.

2. The midpoint of $\overline{PQ}$ is $M_{PQ}(0,4)$

3. Calculate the length of $| \overline{PQ}| = 4\sqrt{5}$

4. Calculate the length of the radius using Pythagorean theorem $r = 2\sqrt{10}$

5. Calculate the length of $|\overline{M_1, M_{PQ}}| = \sqrt{20}$ and $|\overline{M_1, P}| = r$

6. Solve these two equations for $m_{11}, m_{12}$

For confirmation only:
• Jan 12th 2011, 01:10 PM
Soroban
Hello, Beggarsbelief!

Quote:

$\text{Two equal circles intersect at }P(2,0)\text{ and }Q(-2,8).$

$\text{The distance from the centre of each circle}$
. . $\text{to the common chord }PQ\text{ is }2\sqrt{5}$

$\text{Find the equations of the two circles.}$
Code:

                    |                     |  * * *                     *          *                   * |            *                 *  |              *                     |               Q*  |    A        *               * o * |    o        *           *    *  | *            *         *          |  *       *        *  oM  *        *                     |            *       *            *    *    *   ----*---------o---+---*-o-*---------       *        B  |    *P                     |       *            |    *         *          |  *           *        | *               * * * |                     |

The centers of the circles are $\,A$ and $\,B.$
The circles intersect at $P(2,0)$ and $Q(-2,8).$

Draw chord $\,PQ.$
Draw segment $\,AB.$
It can be shown that they intersect at $M(0,4).$

Code:

          Q    |           o    |           *    |          A             *  |          o             *  |        *  :               * |    *    :2           4    *|M *        :     + - - - - - o - - - - - +     :        *  |*    4   2:    *    | *     :  *      4|  *     o          |  *     B          |    * ----------------+-----o--------                 |  2  P                 |

Note that: . $AB \perp PQ$ . . . Quadrilateral $APBQ$ is a rhombus.

We are told that: . $AM = BM = 2\sqrt{5}$
. . .We find that: . $PM = QN = 2\sqrt{5}$ . . . $APBQ$ is a square.

To go from from $\,P$ to $\,M$, we go left 2, up 4.
. . Then, to go from $\,M$ to $\,A$, we got right 4, up 2.
. . Hence, center $\,A$ is at $(4,6).$

. . To go from $\,M$ to $\,B$, we go left 4, down 2.
. . Hence, center $\,B$ is at $(-4,2).$

Now you can write the equations of the circles.

• Jan 12th 2011, 01:21 PM
Quote:

Originally Posted by Beggarsbelief
Two circles intersect at p(2,0) and q(-2,8). The distance from the centre of each circle to the common chord [pq] is √20

Find the equations of the two circles.

That is the question, I have completed though it took me roughly 40 minutes to complete. I figured out the radius through simple triangle geometry and then created 3 equations in g, f, and c (G being the centre of the circle's x value) (F being the centre of the circle's y value) and c being c.

What resulted was a long drawn out form of substitution which took up one full a4 page to write out and solve, substituting in large equations for f, f^2, c, and c^2 which resulted in getting an equation all in g which eventually led me to a result for g and from there on, substituting back in was easy.

My problem is that it took so long, I am wondering is there another way to solve this question, through trigonometry or did I over complicate it do you think? I am sitting my exams in June and I am expected to answer at least 3 questions of this difficulty along with various other varying difficulty questions, do you think there is an easier way to solve this one than the long, drawn out, painful process I endured, or was what I did correct?

Here is another way.

(1) Find the midpoint of the common chord.

(2) Write the equation of the line passing through this midpoint which contains the centre of both circles.

This allows you to write the "y centre co-ordinate" in terms of the x one
or vice versa.

(3) Use the perpendicular distance from a point to a line formula $d=\displaystyle\ |\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}|$
(distance from centres to common chord).

The slope of the common chord is $\displaystyle\frac{8-0}{-2-2}=-2$

The equation of the common chord is $y-0=-2(x-2)\Rightarrow\ 2x+y-4=0$

The midpoint of the common chord is $(0,\;4)$

The equation of the common centreline is $y-4=\frac{1}{2}(x-0)\Rightarrow\ 2y-8=x$

Distance from a point to a line:

$\displaystyle\sqrt{20}=|\frac{2(2y-8)+1(y)-4}{\sqrt{2^2+1^2}}|$

Solving this obtains the two y co-ordinates of the centres
(and subsequently the x co-ordinates).
• Jan 14th 2011, 10:41 AM
Beggarsbelief
Ah I understand what each one of you did. Too bad I didn't think of any of them, these methods are a lot more simpler than the method I tried. Thank you all very much. :)