1. ## Circle Time

Help me to solve this question please~

Q: Find the standard equation of the circle [which means in the form of $(x-a)^2+(y-b)^2=r^2$] that passing through the point $(-3,1)$ and containing the points of intersection of the circles $x^2+y^2+5x=1$ and $x^2+y^2+y=7$.

When i looking for the intersection point of the 2 circles, i found that both of their (x,y) coordinate are irrational numbers (I'm forced to use the quadratic formula to solve it). After that, it become more and more complicated in further solving... And the answer came out, WRONG! Congratulation to myself. I wonder did anyone figure out how to solve it? I can't find another point to form the two bisectors of their chords, in order to get the coordinate of the center, I need to find the intersection point of the two bisectors. So anyone please help me!

2. Can you show how you found the intersection points?

3. To look for the intersection points of two circle, just let them to be equal.
Equation 1:
$x^2+y^2+5x=1$
$x^2+y^2+5x-1=0$
Equation 2:

$x^2+y^2+y=7$
$x^2+y^2+y-7=0$
Eq.1 = Eq.2 :

$x^2+y^2+5x-1=x^2+y^2+y-7$

...
$y=5x+6$

(This is the line equation that from by the two intersection point of the two circle.)

So, substitute back to the circle equation to find out which two points does it intersect with the line.
So, take any one of the circle to substitute in it.
$x^2+y^2+5x-1=0$
$x^2+(5x+6)^2+5x-1=0$ (Replace $y$ by $5x+6$)
$26x^2+65x+35=0$
...
Then is the nightmare - The Irrational Number.
Can you solve it without using the Annoying-Irra-XY-Coordinates?

4. This work looks good to me. So you should have three points on a circle and thus can solve for the radius. Is that where you're getting tripped up? I checked over your work trying to spot for errors but it looks correct so far.

5. Jameson, is that mean that the Irrational Numbers for the coordinates are unavoidable? And there's no alternate way to solve it?

6. ## correct circle intersection equation?

hmmm, I don't think you can just set the two circle equations to be equal :
$x^2+y^2+5x-1=x^2+y^2+y-7$

I think what you're supposed to do is solve the system made up of both equations. To be sure you should try to get the original circle equations :
$x^2+y^2+5x-1=0$
$(x+5/2)^2+y^2-21/4=0$
and
$x^2+y^2+y-7=0$
$x^2+(y+1/2)^2-27/4=0$

so one circle is centered at $(-5/2,0)$ with a ray of $sqrt(21)/2$ and the other is at $(0, -1/2)$ with ray $sqrt(27)/2$... Unless I'm mistaken, that looks lie a guarantee that the results will be irrational. I sketched the circles and the intersections seem to invalidate the $y=5x+6$ line, then again my sketches are terrible.

hope this helps

7. Originally Posted by kaze_windsa
Jameson, is that mean that the Irrational Numbers for the coordinates are unavoidable? And there's no alternate way to solve it?
I've made a sketch (to be honest: My program made a sketch).

At the left side you find all approximate solutions.

8. Looks like your line equation was right actually!

9. Originally Posted by kaze_windsa
...Then is the nightmare - The Irrational Number.
Can you solve it without using the Annoying-Irra-XY-Coordinates?
No.

All your considerations and calculations are OK.

Refering to my previous post you'll get:

$A\left(-\dfrac54-\dfrac3{52} \sqrt{65}\ ,\ -\dfrac14-\dfrac{15}{52} \sqrt{65} \right)$ ...... $B\left(-\dfrac54+\dfrac3{52} \sqrt{65}\ ,\ -\dfrac14 + \dfrac{15}{52} \sqrt{65} \right)$

When you have calculated the x-values plug them into the equation of the line to get the y-coordinates.

10. As the others have confirmed, the intersections are unfortunately irrational. If it helps, you should get $\left(\frac{-3\sqrt{65}-65}{52},\,\frac{-15\sqrt{65}-13}{52}\right)$ and $\left(\frac{3\sqrt{65}-65}{52},\,\frac{15\sqrt{65}-13}{52}\right)$ for the points after solving the quadratic. Finding the equation of the circle may not be easy to do by hand, but it is certainly possible.

11. Please note that the question asked about general intersections has been moved to its own thread. It is a good question and deserves its own place.

12. kaze - now that you have your intersection points, are you able to construct the final solution?

13. Thanks Jameson.

Now I try to use the Irrational-Number's Coordinate back to continue solving again.

My Method of Solving (continue):
(Use the 3 points on the circumference and form 2 chords{one of them will be used both of the time}, then find their bisectors. Finally find their intersection point again in order to find the center. Then find its radius. Volla!)

Fact(?):
(However, last time just I used the same method as told here, but the answer is wrong, because my tutor said that the coordinate of center of the circle is a rational numbers, both $x$ and $y$ of them, but what i got was a mess-numbers. Maybe just because of my typo error on my calculator.)

I will post my answer as soon as possible when I get my answer, after get checked by my tutor.

Thanks for anyone who guide me all the way. Much appreciated.