Yes, I think...
First mark O and T the centers...
Then mark as M the mid of PQ.
Find PO and PT
Then use the formula between two points...
hmmm... (sorry for the mess... 1 minute thinking.)
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?
1. Draw a sketch.
Let and denote the midpoints of the circles and respectively.
2. The midpoint of is
3. Calculate the length of
4. Calculate the length of the radius using Pythagorean theorem
5. Calculate the length of and
6. Solve these two equations for
For confirmation only:
Code:| | * * * * * * | * * | * | Q* | A * * o * | o * * * | * * * | * * * oM * * | * * * * * ----*---------o---+---*-o-*--------- * B | *P | * | * * | * * | * * * * | |
The centers of the circles are and
The circles intersect at and
It can be shown that they intersect at
Code:Q | o | * | A * | o * | * : * | * :2 4 *|M * : + - - - - - o - - - - - + : * |* 4 2: * | * : * 4| * o | * B | * ----------------+-----o-------- | 2 P |
Note that: . . . . Quadrilateral is a rhombus.
We are told that: .
. . .We find that: . . . . is a square.
To go from from to , we go left 2, up 4.
. . Then, to go from to , we got right 4, up 2.
. . Hence, center is at
. . To go from to , we go left 4, down 2.
. . Hence, center is at
Now you can write the equations of the circles.
(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
(distance from centres to common chord).
The slope of the common chord is
The equation of the common chord is
The midpoint of the common chord is
The equation of the common centreline is
Distance from a point to a line:
Solving this obtains the two y co-ordinates of the centres
(and subsequently the x co-ordinates).