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Thread: Circle/linear intersection proof

  1. #1
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    Circle/linear intersection proof

    Hi all,

    Prove that, for any value of c, the circle whose equation is $\displaystyle (x+c)^2+(y-c)^2=c^2$ touches both the x-axis and the y-axis.

    Is it enough here to show that $\displaystyle b^2-4ac=0$ when either linear equation $\displaystyle x=0, y=0$ is plugged into the the circle equation.

    discriminant in both cases is $\displaystyle 4c^2-4c^2=0$

    or is this incorrect?

    thanks sammy
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  2. #2
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    The Equation circle centered at (m, n) wibth radius R is :

    $\displaystyle (x-m)^2+(y-n)^2 = R^2$

    Since comparative data circuit is :

    $\displaystyle (x+c)^2+(y-c)^2 = c^2$

    After all, this is a circle centered at the point (c, c-) with radius | c |.



    I think that this trick question ... To know that the radius must always be positive that this is a definition (c tells us that real, not a given that the charges). For c = 0 of course it is a circle centered at the origin and radius 0, ie a point so it meets the conditions are empty. Any c different from 0 (positive or negative) point (c, c-) of course exists, then there is only that | c | then there is always such a positive cycle
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  3. #3
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    Hello, sammy28!

    Prove that, for any value of $\displaystyle c$, the circle $\displaystyle (x+c)^2+(y-c)^2\:=\:c^2$
    touches both the $\displaystyle x$-axis and the $\displaystyle y$-axis.

    Let $\displaystyle y = 0\!:\;\;(x+c)^2 + (-c)^2 \:=\:c^2 \quad\Rightarrow\quad (x+c)^2 \:=\:0$

    . . . . . . . . . . $\displaystyle x+c \:=\:0 \quad\Rightarrow\quad x \:=\:-c$

    The circle has one $\displaystyle x$-intercept; it is tangent to the $\displaystyle x$-axis.



    Let $\displaystyle x = 0\!:\;\;(c)^2 + (y-c)^2 \:=\:c^2 \quad\Rightarrow\quad (y-c)^2 \:=\:0$

    . . . . . . . . . . $\displaystyle y-c \:=\:0 \quad\Rightarrow\quad y \:=\:c$

    The circle has one $\displaystyle y$-intercept; it is tangent to the $\displaystyle y$-axis.



    Edit: I corrected my typos.
    .
    Last edited by Soroban; Jul 28th 2010 at 06:53 AM.
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  4. #4
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    Quote Originally Posted by Soroban View Post

    Let $\displaystyle x = 0\!:\;\;(c)^2 + (x-c)^2 \:=\:c^2 \quad\Rightarrow\quad (x-c)^2 \:=\:0$

    . . . . . . . . . . $\displaystyle x-c \:=\:0 \quad\Rightarrow\quad x \:=\:c$

    The circle has one $\displaystyle y$-intercept; it is tangent to the $\displaystyle y$-axis.
    thanks yehoram and soroban . I like soroban's answer since it states the exact equations for all values of c.

    i think the above should be

    Let $\displaystyle x = 0\!:\;\;(c)^2 + (y-c)^2 \:=\:c^2 \quad\Rightarrow\quad (y-c)^2 \:=\:0$

    . . . . . . . . . . $\displaystyle y-c \:=\:0 \quad\Rightarrow\quad y \:=\:c$

    The circle has one $\displaystyle y$-intercept; it is tangent to the $\displaystyle y$-axis.
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