Question

How do u solve {x^2+y^2=63
x^2-3y^2=27

Answer 1

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

Answer 2

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

and the second eqn. is a hyperbola

Answer 3

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

Answer 4

let's put the hyperbola in the standard form

\[{x^2 \over 27} - {y^2 \over 9} =1\]

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

anyway, now we will start solving this algebraically

Answer 10

the circle eqn can give us\[x^2 = 63-y^2\]

Answer 11

so if I substitute this into the hyperbola eqn

\[(63 - y^2) - 3y^2 = 27\]

Answer 12

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

Answer 13

since \[-4y^2 = -44 \]

\[y = \pm 4\]

Answer 14

if you plug this into the circle,

\[x^2 + (\pm 4)^2 = 63\]

implies that

\[x = \pm \sqrt{47}\]

Answer 15

unless my calculations are wrong,
the four points that the circle and the hyperbola intersect are

\[(\pm \sqrt{47},\pm 4)\]

Answer 16

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

Answer 17

\[\left\{x\to -3 \sqrt{6},y\to -3\right\},\left\{x\to -3 \sqrt{6},y\to 3\right\},\left\{x\to 3 \sqrt{6},y\to -3\right\},\left\{x\to 3 \sqrt{6},y\to 3\right\} \]
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.