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Math Help - Equation of Circle Tangent

  1. #1
    Member Henderson's Avatar
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    Equation of Circle Tangent

    To begin with, I solved this problem, and got the correct answer. The reason I'm posting is because the solution manual uses a different method, and I can't follow what they're doing. Please explain their solution.

    Problem:
    For what values of k is the straight line y = kx + 1 a tangent to the circle with center (5,1) and radius 3?

    My solution:
    The distance from the y-intercept (0,1) to the center of the circle is 5, and the radius is 3. Since the tangent is perpendicular to the radius, a 3-4-5 triangle is created. Dropping an altitude in this triangle gives us a change-in-y, change-in-x triangle for the tangent line. Since it is similar to the large triangle, it also has an opposite/adjacent ratio of \frac{3}{4}, which is the slope of our tangent line. Similarly, there will be a symmetric tangent line with a slope of \frac{-3}{4}.

    Their solution:

     <br />
\frac{|5k-1+1|}{\sqrt{k^2+1}} = 3<br />

    ...(algebraic work that I can follow)...

     <br />
k = \pm\frac{3}{4}<br />

    Where is their starting equation coming from? Is this some distance from a line to a point equation that I've never seen?
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  2. #2
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    Quote Originally Posted by Henderson View Post
    Problem:
    For what values of k is the straight line y = kx + 1 a tangent to the circle with center (5,1) and radius 3?
    Their solution:
     <br />
\frac{|5k-1+1|}{\sqrt{k^2+1}} = 3<br />
    ...(algebraic work that I can follow)...
     <br />
k = \pm\frac{3}{4}<br />

    Where is their starting equation coming from? Is this some distance from a line to a point equation that I've never seen?
    The distance from the point (p,q) to the line Ax+By+C=0 is given as:
    \dfrac{|Ap+Bq+C|}{\sqrt{A^2+B^2}}
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  3. #3
    Member Henderson's Avatar
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    Thanks- that's definitely the equation they're using. Can you point me toward a derivation of that formula? I've never seen it before.
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  4. #4
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    Quote Originally Posted by Henderson View Post
    Thanks- that's definitely the equation they're using. Can you point me toward a derivation of that formula? I've never seen it before.
    As you can see this best done using vectors.
    Thanks from Henderson
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