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    Show that the lengths of the tangents from the point (h, k) to the circle X^2 + Y^2 + 2fX + 2gY = 0 are the square root of (h^2 +K^2 + 2fh + 2gk +c ) ^1/2. If anyone could please help. I'd be sincerely grateful. Thanks. van.domo
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    (Original post by van.domo)
    Show that the lengths of the tangents from the point (h, k) to the circle X^2 + Y^2 + 2fX + 2gY = 0 are the square root of (h^2 +K^2 + 2fh + 2gk +c ) ^1/2. If anyone could please help. I'd be sincerely grateful. Thanks. van.domo
    There may be a typo in your question ... but the general idea would be to get the equation of the circle into the usual form to get the centre and radius.

    Now work out the distance from (h,k) to the centre. Draw a diagram showing the tangent and radius at that point ....
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    (Original post by van.domo)
    Show that the lengths of the tangents from the point (h, k) to the circle X^2 + Y^2 + 2fX + 2gY = 0 are the square root of (h^2 +K^2 + 2fh + 2gk +c ) ^1/2. If anyone could please help. I'd be sincerely grateful. Thanks. van.domo
    Le t point P(h,k)
    The line passing through the midpoint of the circle intersects the circle
    at point A and B, then PA*PB=e^2
    where e is the length of the tangent to the circle from P.
    This is the theorem of power of P relating to (or on) the circle
    The circle:
    (x+f)^2+(y+g)^2=f^2+g^2
    So the midpoint of the circle is O(-f,-g)
    and the radius is \sqrt{f^2+g^2}
    PA=PO-r
    PB=PO+r
    PO=\sqrt{(h+f)^2+(k+g)^2}
    so
    e^2=PA\cdot PB=(PO-r)(PO+r)=PO^2-r^2
    Substitute back and arrange
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    You know. It's seems a hard one for C1. Thanks very much for your help and support. It's appreciated. Domo
 
 
 
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