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Thread: possible value

  1. #1
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    possible value

    hi, my question is:

    find the possible value of the constant c so that the line $\displaystyle y = 2x + c$ is a tangent to the circle $\displaystyle x^2 + y^2 = 4$

    i treated it as a simultaneous equation (if thats the right thing to do) and continued by making equation one $\displaystyle y^2 = (2x + c)^2$ and subbed it into the second like this $\displaystyle x^2 + (2x + c)^2 = 4$ then expanded the second to $\displaystyle x^2 + 4x^2 + 4xc + c^2 = 4$ then simplified to $\displaystyle 5x^2 + 4xc + c^2 = 4$

    i don't know what to do from here, can someone help me out please? thanks
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  2. #2
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    Hello mark
    Quote Originally Posted by mark View Post
    hi, my question is:

    find the possible value of the constant c so that the line $\displaystyle y = 2x + c$ is a tangent to the circle $\displaystyle x^2 + y^2 = 4$

    i treated it as a simultaneous equation (if thats the right thing to do) and continued by making equation one $\displaystyle y^2 = (2x + c)^2$ and subbed it into the second like this $\displaystyle x^2 + (2x + c)^2 = 4$ then expanded the second to $\displaystyle x^2 + 4x^2 + 4xc + c^2 = 4$ then simplified to $\displaystyle 5x^2 + 4xc + c^2 = 4$

    i don't know what to do from here, can someone help me out please? thanks
    Thanks for showing us your working so far. You are quite correct in what you've done. The next step is to re-arrange as a quadratic in $\displaystyle x$:

    $\displaystyle 5x^2 +4c\cdot x+c^2-4=0$

    and then say that the line is a tangent if the two points of intersection coincide at one point; i.e. if this quadratic has equal roots. So, use $\displaystyle b^2-4ac=0$, and solve for $\displaystyle c$.

    (I make the answers $\displaystyle c = \pm 2\sqrt5$.)

    Grandad
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  3. #3
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    thanks for answering but i don't think i understand what you mean by rearrange as a quadratic in x? why have you set it out like this $\displaystyle 5x^2 +4c\cdot x+c^2-4=0$?

    if i was using $\displaystyle \sqrt{b^2 - 4ac} = 0$ i would have made it $\displaystyle \sqrt{16x^2c^2 - 4(5)(-4)} = 0$
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  4. #4
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    Hello mark
    Quote Originally Posted by mark View Post
    thanks for answering but i don't think i understand what you mean by rearrange as a quadratic in x? why have you set it out like this $\displaystyle 5x^2 +4c\cdot x+c^2-4=0$?

    if i was using $\displaystyle \sqrt{b^2 - 4ac} = 0$ i would have made it $\displaystyle \sqrt{16x^2c^2 - 4(5)(-4)} = 0$
    I mean re-arrange to form an equation of the form $\displaystyle ax^2 + bx + c=0$.

    This equation has equal roots if $\displaystyle b^2-4ac=0$.

    So in the equation $\displaystyle 5x^2 +4c\cdot x+c^2-4=0$

    • $\displaystyle a=5$
    • $\displaystyle b=4c$
    • $\displaystyle c=(c^2-4)$

    So $\displaystyle b^2 - 4ac = (4c)^2 - 4\cdot5\cdot(c^2-4) = 16c^2-20(c^2-4)=80-4c^2$

    So $\displaystyle b^2-4ac = 0 \Rightarrow c^2=20 \Rightarrow c = \pm2\sqrt5$

    Grandad
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  5. #5
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    ah i understand now, thanks for helping
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