# Intersecting lines,circles,and parabolas

1. Sep 24, 2011

### Sumaya

1. The problem statement, all variables and given/known data

find the points in which the graphs intersect

2. Relevant equations

y=2x, x^2+y^2=1

3. The attempt at a solution

the center point of the circle is (0,0)
and i collect some points to draw the line equation
(-1,-2) (0,0) (1,2) (2,4) etc ..

but i don't know how to get the intersect points ..
and also for the parabola equation ... are they have same way to know the intersect points ...
is there a rule or what ??
please explain to me in simple english words..

thanx a lot ...

2. Sep 24, 2011

### eumyang

Just solve for x and y. Plug y = 2x into
x2+y2=1
to get
x2+(2x)2=1

Solve for x (you should get two solutions), and plug both into
y = 2x
to get the corresponding solutions for y.

The 2nd equation is an equation of a circle, not a parabola.

3. Sep 24, 2011

### HallsofIvy

In simple English (nicht Deutsch, warum?), don't worry about the geometry, solve the equations!

If the problem were to find the intersection of $x^2+ y^2= 1$ and $y= x^2$, the first is a circle and the second a parabola. That's nice to know (it tells us we can expect to find two points of intersection) but not necessary to the solution. Since $y= x^2$ we can replace $x^2$ by y in the first equation: $y+ y^2= 1$ or $y^2+ y- 1= 0$. That's a quadratic equation and we can either complete the square or use the quadratic formula to solve for y. Once we have found y, x is a square root.

(The quadratic equation has two roots, of course, and you might think that since each has two roots, there would be 4 (x, y) combinations. But one of the (real) roots to the quadratic is negative. That gives only imaginary roots and coordinates of points in a graph must be real. Only y> 0 gives the two points of intersection.)

4. Sep 26, 2011

### Sumaya

thanx alot ...

5. Sep 26, 2011