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Math Help - Really tough circle problem

  1. #1
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    Really tough circle problem

    Hi, My problem is this : find the equation of the circle determined by the given conditions :

    center (-2,3) and tangent to the line 4y-3x+2=0.

    the equation for the line is y=3/4x -1/2 (im pretty sure)

    so I drew that line, then I figured I would need to use the distance formula from where the line starting at (-2,3) with slope -4/3 (negative reciprical of 3/4) intersects 4y-3x+2=0.

    However, it doesnt intersect at a point, its an arbitrary point. I cant determine the exact x,y of it by graphing. I dont see how to make the distance formula = to an arbitrary point. Also the only other thing I can think of is that the circle can only intersect 4y-3x+2=0 once not twice. I also cant figure out how to make any triangles that would help me. Im stumped. What Do I do? Who knows the answer? I want to know badly.
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  2. #2
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    You almost have it. *Really* close.

    Formula for line from (-2, 3) with slope -4/3 is
    y - 3 = -4/3(x + 2)

    Now find where this intersects
    4y - 3x + 2 = 0

    These should intersect at one point, P, because they are perpendicular.

    Calculate the distance from (-2, 3) to P, and this is the radius for your circle.

    This touches the line once. Why? The radius from (-2,3) to P forms right angles with the line.
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  3. #3
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    The distance from (-2,3) to the line 3x-4y-2=0 is \dfrac{|3(-2)-4(3)-2|}{\sqrt{(3)^2+(-4)^2}}.
    That distance is the radius of the circle.
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  4. #4
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    I want to know how to find p. When I graphed it P was not a point I could see, it was between different coordinates. How can I get the distance between the center and P if I have to guess what P is because its not on an exact point (meaning x and y are fractions and I cant tell what the fractions are.)

    Also I got another similar question. - Center on the line x+y=1, Passes through (-2,1) and (-4,3):

    I think that -2,1 and -4,3 should be focus and directx of a parabola, and any point on that parabola that hits the line x+y=1 should be the center since it will be equal distance from the 2 points and also on the line.

    I got (y-2)=1/4(x+3) as the equation of the parabola (used midpoint formula to get vertex) and y=-x+1 as the line.

    but I have the same problem as the other question. The line intersects the parabola at a point I can not tell when I graph them. x and y are fractions not exact points. How do I figure out problems that bottle down like this?

    I need to know how to find P basicly. I mean, its possible my math was just wrong and P turned out to be a point on the graph though that is actual numbers not fractions.

    (also how do I thank people who try to help me? I see it has a thanked option)
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  5. #5
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    When two lines intersect, their x and y coordinates are equal.

    y - 3 = -4/3(x + 2) intersects 4y - 3x + 2 = 0

    at the x and y that simultaneously solves the 2 equations.
    This solution for x and y is P.

    Also you can use the formula for distance from point to line formula provided in an earlier post (without directly finding P).
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  6. #6
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    To answer the OP, all one needs is the distance from the center to the tangent line.
    We do not need to find p.
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  7. #7
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    Your parabola question is a bit unclear.
    Did you mean (y-2) = 1/4(x+3)^2 ?

    To find the intersection between the parabola and the line y=-x+1

    Again, you find the simultaneous solution(s) for x and y that solve both
    (y-2) = 1/4(x+3)^2 and y=-x+1

    Example: You can use substitution to get
    (y-2) = (-x+1-2) = -x-1 = 1/4(x+3)^2
    solve for x
    and then use this to find y = -x+1
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  8. #8
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    yeah i meant ^2 on the end of that parabola.

    Also, I realize that the answer is the distance between p and center.

    By graphing the line 4y-3x+2=0 I can not clearly tell which point intersects with the perpendicular line from the center of the circle. Therefore how can I tell what to input in the distance formula? This is what I want to know.

    OK to best clear this up in my mind, You can tell me what to fill into this equation :

    (x - -2)^2 + (y - 3)^2 = Z

    What is x and y and how did you get x and y.

    (-2 and 3 are the center of the circle, x and y are the points I cant clearly see when i graph the equation 4y-3x+2=0. also I understand what point it should be (the intersection) I just cant tell what it is)
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  9. #9
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    Alright, lets do it by the intersection method.

    y - 3 = -4/3(x + 2) intersects 4y - 3x + 2 = 0 where?
    Lets solve this system (substitue y = -4/3(x+2) + 3 into the second equation)
    4(-4/3(x+2) + 3) - 3x + 2 = 0
    Solving give x = 2/5
    y = -4/3(2/5 + 2) + 3 = -1/5

    so P = (2/5, -1/5)

    Distance from (-2,3) to (2/5,-1/5)?
    sqrt((-2 - 2/5)^2 + (3 + 1/5)^2) = 4

    This means the radius of your circle is 4.

    Equation for the circle at center -2, 3 with radius 4 is
    (x + 2)^2 + (y - 3)^2 = 4^2
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  10. #10
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    Quote Originally Posted by snowtea View Post
    Your parabola question is a bit unclear.
    Did you mean (y-2) = 1/4(x+3)^2 ?

    To find the intersection between the parabola and the line y=-x+1

    Again, you find the simultaneous solution(s) for x and y that solve both
    (y-2) = 1/4(x+3)^2 and y=-x+1

    Example: You can use substitution to get
    (y-2) = (-x+1-2) = -x-1 = 1/4(x+3)^2
    solve for x
    and then use this to find y = -x+1
    hold on sorry didnt see this post trying it now. Also plato I saw the formula you posted but I didnt get why it would work. never seen that before.
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  11. #11
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    On a tangent line to a circle, the point of tangency is the nearest point to the center.
    So the distance from the center to a tangent line is the radius of the circle.
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  12. #12
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    Awsome ! I never thought to substitute. Oops. I guess that when you substitute it takes the 1st equation and just adds the parameters of the 2nd. Then when you solve bang. Cool I so get it now. thank you so much to both of you for helping!


    Ps how do I thank you guys? I see you get creds for helping people.
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  13. #13
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    Quote Originally Posted by Plato View Post
    The distance from (-2,3) to the line 3x-4y-2=0 is \dfrac{|3(-2)-4(3)-2|}{\sqrt{(3)^2+(-4)^2}}.
    That distance is the radius of the circle.
    Why does dividing the absolute value of the center into the equation of the line (without the =0) by the square root of the constants to the line squared give you the distance of the radius of the circle? How you help me picture why that works ?
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  14. #14
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    Glad you got it. I think you were already really close from the beginning.

    You can thank people by clicking the thanks button (thumbs up icon) under the post.
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  15. #15
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    The formula is a shortcut for what you did with the perpendicular line.
    If you solve for P algebraically and calculate this distance, it simplifies to the formula.
    You can derive the formula in this way.

    The reason why the equation looks so simple requires understanding a bit about vectors and dot products. You don't have to worry about it if you haven't learned about vectors yet.
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