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Thread: Find the best way to solve this question

  1. #1
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    Find the best way to solve this question

    Find the equation of the circle which is tangent to the line $\displaystyle 3x - y - 1 = 0 ~ $ at $\displaystyle A(1,2)~$ and passes through the point $\displaystyle B(5,-2)$

    I've got the best way to solve this problem , do you know how i solved it ?
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Well, I did the usual textbook stuff

    Spoiler:
    First I found the equation of the line normal to $\displaystyle 3x-y-1=0$ and passing through A. The centre of the circle lies on this line. Then I found the equation of the perpendicular bisector of the chord AB. The centre of the circle also lies on this line. Solving for the point of intersection, I determined the centre of the circle to be $\displaystyle (4,1).$ Finally, I found the radius of the circle, which is $\displaystyle \sqrt{10}.$ Hence the circle is $\displaystyle (x-4)^2+(y-1)^2=10.$

    Nevertheless the method did not appear to be particularly quick, so I guess you have a quicker solution.
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  3. #3
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    Let F be a family of circles such that all members of the family F touch the line $\displaystyle 3x - y - 1 = 0 ~~$at $\displaystyle A(1,2)$ .

    Then F should be $\displaystyle (x-1)^2 + (y-2)^2 + k(3x - y - 1) = 0$ .

    Since the required member passes through $\displaystyle B(5,-2)$ ,

    it satisifies $\displaystyle (5-1)^2 + (-2-2)^2 + k[3(5) - (-2) - 1] = 0$ .

    Therefore , $\displaystyle k = -2 $

    Finally substitute back to F ,

    The required equation is :

    $\displaystyle (x-1)^2 + (y-2)^2 -2(3x-y-1) = 0$
    $\displaystyle x^2 + y^2 - 8x - 2y + 7 = 0$
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