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anonymous

  • 5 years ago

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?

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  1. amistre64
    • 5 years ago
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    circles arent normally linear...

  2. amistre64
    • 5 years ago
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    those may be more ovalish if anything, but same principle :)

  3. anonymous
    • 5 years ago
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    So its right though?

  4. amistre64
    • 5 years ago
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    cant tell, im still confused by the question :)

  5. anonymous
    • 5 years ago
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    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.

  6. amistre64
    • 5 years ago
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    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

  7. amistre64
    • 5 years ago
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    your best bet would be to "eliminate" a variable first

  8. amistre64
    • 5 years ago
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    (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 :)

  9. amistre64
    • 5 years ago
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    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))

  10. anonymous
    • 5 years ago
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    Thank you!

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