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Thread: Condition for a chord of contact to subtend a right angle at the center.

  1. #1
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    Condition for a chord of contact to subtend a right angle at the center.

    This is a question form SL Loney's The Elements of Coordinate Geometry :

    Find the condition that the chord of contact of tangents from the point (x' , y') to the circle $\displaystyle x^2 + y^2 = a^2 $ should subtend a right angle at the center.

    Now what I can propose is, the equation to the chord of contact is $\displaystyle xx' + yy' = a^2 $ and that of circle is $\displaystyle x^2 + y^2 = a^2 $ so we can solve this system of equations and the values that we would get is the end points of chord and hence we could find the length of the chord AB. Now to subtend a right angle it must follow the Pythagorean theorem i.e. $\displaystyle a^2 + a^2 = AB^2 $. I have added an attachment to illustrate myself more clearly.

    But in this approach I drown into unsolvable algebra and can't get further. So, I found a solution of it (I have added it in attachment ), but in the solution I don't understand when the author says "we make (1) homogenous with the help of (2)". My doubts are :-

    1. What is homogenous after all?
    2. Why do we need to do that? Can't we solve it by the principles which I have stated above?
    3. Has the author done it only for the sake of solving that particular problem ?

    Thank you. Any help will be much appreciated.
    Attached Thumbnails Attached Thumbnails Condition for a chord of contact to subtend a right angle at the center.-sl-loney.png   Condition for a chord of contact to subtend a right angle at the center.-my-math.png  
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  2. #2
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    Re: Condition for a chord of contact to subtend a right angle at the center.

    $\displaystyle OAPB $ is a square therefore

    $\displaystyle (x')^2+(y')^2=\text{OP}^2=\text{AB}^2=2a^2$
    Last edited by Idea; Aug 17th 2019 at 10:47 PM.
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