Question

Set both equations to y=mx+b
x^2+y^2=17
x^2+y^2-2x=13

(sqrt-x^2 + 17 ) (-sqrt-x^2+17)
(sqrt-x^2+2x+13) (-sqrt-x^2+2x+13)
Is this right?

Answer 1

circles arent normally linear...

Answer 2

those may be more ovalish if anything, but same principle :)

Answer 3

So its right though?

Answer 4

cant tell, im still confused by the question :)

Answer 5

im supposed to put the equations in mx+b so I can put it in my graphing calculator. Then im supposed to use the intersect function on my calculator to find the points. there should be 4 points.

Answer 6

there are up to 4 possible points. that is right...

but:

y=mx+b is the equation for a line.

What you have here are 2 ovals that might be intersecting adn there is no way to make any part of them a straight line that I am aware of

Answer 7

your best bet would be to "eliminate" a variable first

Answer 8

(x^2+y^2-0x=17) (-1)
x^2+y^2-2x =13

-x^2 -y^2 +0x =-17
x^2 +y^2 -2x = 13
--------------------
0x^2 +0y^2 -2x = -4

-2x = -4
x = 2

plug in x=2 for your x values and solve for y :)

Answer 9

4 + y^2 = 17
y^2 = 13

y = sqrt(13) or y=-sqrt(13)

4 +y^2 -4 = 13
y^2 = 13

y= sqrt(13) or y=-sqrt(13)

your points of intersection are:

(2,sqrt(13)) and (2,-sqrt(13))

Answer 10

Thank you!