## anonymous 5 years ago How do u solve {x^2+y^2=63 x^2-3y^2=27

1. anonymous

there are a couple of ways to solve this, but the most efficient way is to roughly see how the graph is and find out how many intersections there are going to be

2. anonymous

the first eqn. is a circle with radius sqrt(63) and the second eqn. is a hyperbola

3. anonymous

as you can easily see , both of the conics have the center located at the origin, so the number of intersection points can vary from none to 4

4. anonymous

let's put the hyperbola in the standard form ${x^2 \over 27} - {y^2 \over 9} =1$

5. anonymous

from here we can observe that this hyperbola opens sideways, and the verteces are located sqrt(27) units right and left, 3 units up and down

6. anonymous

oops, my bad, only sqrt(27) units right to left. the 3 units up and down tells us the asymptotes so never mind about that

7. anonymous

since the verteces are located sqrt(27) units from the origin, and the circle has a radius of sqrt(63) we can see that the hyperbola crosses the circle four times

8. anonymous

knowing this is actually important, because you don't want to waste time solving the system of eqn. only to find out there were no solutions.

9. anonymous

anyway, now we will start solving this algebraically

10. anonymous

the circle eqn can give us$x^2 = 63-y^2$

11. anonymous

so if I substitute this into the hyperbola eqn $(63 - y^2) - 3y^2 = 27$

12. anonymous

as you can see, everything in the eqn is in terms of y, so we can solve it without any trouble

13. anonymous

since $-4y^2 = -44$ $y = \pm 4$

14. anonymous

if you plug this into the circle, $x^2 + (\pm 4)^2 = 63$ implies that $x = \pm \sqrt{47}$

15. anonymous

unless my calculations are wrong, the four points that the circle and the hyperbola intersect are $(\pm \sqrt{47},\pm 4)$

16. anonymous

wow, I did all this and skull is not even online :( oh well, let me know if anyone found this useful :)

17. anonymous

$\left\{x\to -3 \sqrt{6},y\to -3\right\},\left\{x\to -3 \sqrt{6},y\to 3\right\},\left\{x\to 3 \sqrt{6},y\to -3\right\},\left\{x\to 3 \sqrt{6},y\to 3\right\}$ A plot is attached. Used the following Mathematica statement to solve it $\text{Solve}\left[\left\{x^2+y^2==63,x^2-3 y^2==27\right\},\{x,y\}\right]$ Plot expression follows: Plot[%87, {x, -10, 10}, AspectRatio -> Automatic] where %87 = $\left\{-\sqrt{63-x^2},\sqrt{63-x^2},-\frac{\sqrt{-27+x^2}}{\sqrt{3}},\frac{\sqrt{-27+x^2}}{\sqrt{3}}\right\}$ yuki, Excellent advise with regard to having a plot to look at the solutions.