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Math Help - Equation of a Tangent to a Circle

  1. #1
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    Equation of a Tangent to a Circle

    A Circle with center (-4,2) has a line tangent to it at (-2,6). Find the equation of the tangent.

    I want to check if i got the right answer for this question.

    I found the slope of the line of both points which came out to be 1/2.

    Since the the line is tangent the slope would be the reciprocal of 1/2 which is -2.

    Plug in the slope with the points of the tangent (-2,6)... and I got the equation y=-2x+2.

    Does this seem right? It just looks like it's too easy...
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  2. #2
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    That is great reasoning. I made a mistake in my original post the slope should be 2 just as jhevon has said.
    Last edited by TheEmptySet; February 23rd 2008 at 08:58 PM. Reason: I made a mistake
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chrozer View Post
    A Circle with center (-4,2) has a line tangent to it at (-2,6). Find the equation of the tangent.

    I want to check if i got the right answer for this question.

    I found the slope of the line of both points which came out to be 1/2.

    Since the the line is tangent the slope would be the reciprocal of 1/2 which is -2.

    Plug in the slope with the points of the tangent (-2,6)... and I got the equation y=-2x+2.

    Does this seem right? It just looks like it's too easy...
    your reasoning is correct. but i think you made an error with the slope. the slope of the line connecting the two points is 2 i believe

    ...or maybe i need some sleep. in any case, double check your answer. when all said and done, your line should be y = -(1/2)x + 5
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    your reasoning is correct. but i think you made an error with the slope. the slope of the line connecting the two points is 2 i believe

    ...or maybe i need some sleep. in any case, double check your answer. when all said and done, your line should be y = -(1/2)x + 5
    Actually that equation, which your right about being the line connecting the two points, would be correct, but I have to find the equation of the line going through the points (-2,6) while at the same time being tangent to a circle with center (-4,2).

    So everything is fine and I got the answer right?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chrozer View Post
    Actually that equation, which your right about being the line connecting the two points, would be correct, but I have to find the equation of the line going through the points (-2,6) while at the same time being tangent to a circle with center (-4,2).

    So everything is fine and I got the answer right?
    yes, that's what i meant.

    the slope of the line passing through (-4,2) and (-2,6) is \frac {6 - 2}{-2 - (-4)} = \frac 42 = 2. the tangent line is perpendicular to this line, so the slope of the tangent line is -1/2

    so you want the line with slope -1/2 passing through (-2,6)
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    yes, that's what i meant.

    the slope of the line passing through (-4,2) and (-2,6) is \frac {6 - 2}{-2 - (-4)} = \frac 42 = 2. the tangent line is perpendicular to this line, so the slope of the tangent line is -1/2

    so you want the line with slope -1/2 passing through (-2,6)
    Oh yeah...you're right. I had my it \frac {x - x}{y - y} instead of \frac {y - y}{x - x}
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chrozer View Post
    Oh yeah...you're right. I had my it (x-x)/(y-y) instead of (y-y)/(x-x)
    ok. glad we cleared that up. now you can get the solution

    use [tex]\frac {y_2 - y_1}{x_2 - x_1}[/tex] to get \frac {y_2 - y_1}{x_2 - x_1}
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    ok. glad we cleared that up. now you can get the solution

    use [tex]\frac {y_2 - y_1}{x_2 - x_1}[/tex] to get \frac {y_2 - y_1}{x_2 - x_1}
    Ok...thanks. Now time for me to go to sleep
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