Triangle translated

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The triangle formed by the points ${\rm{A}}\left( { - {\rm{9}},{\rm{ 29}}} \right),{\rm{ B}}\left( {{\rm{1}}5,\,22} \right), {\rm{C}}\left( {{\rm{3}},{\rm{ 13}}} \right)$ is translated parallel to the y-axis so that the sides AC and BC touch the circles ${x^2} + {y^2} + 16x - 2y - 35 = 0$ and ${x^2} + {y^2} - 20x - 4y + 79 = 0$ respectively at $A'$ and $B'$ . Find the perimeter of the quadrilateral formed by $A'$ , $B'$ and the corresponding new positions of A and B.

50 - 5\sqrt 2

25 + 5\sqrt 2

50

50 + 5\sqrt 2

1 solution

Vidyamanohar Sharma
Aug 13, 2015

Observe that triangle ABC is right angled at C. Slope of BC= $\frac{3}{4}$ . So, BC in new position is ${\rm{3x}} - {\rm{4y}} + {\rm{3}} = 0{\rm{ or 3x}} - {\rm{4y}} - {\rm{47}} = 0$ . Nearer to B is ${\rm{3x}} - {\rm{4y}} + {\rm{3}} = 0$ Similarly, Slope of AC= $- \frac{4}{3}$ . So, AC in new position is ${\rm{4x}} + 3{\rm{y}} - {\rm{21}} = 0{\rm{ or 4x}} + 3{\rm{y}} + 79 = 0$ . Nearer to A is ${\rm{4x}} + 3{\rm{y}} - {\rm{21}} = 0$ . Hence the required tangents are ${\rm{3x}} - {\rm{4y}} + {\rm{3}} = 0$ , ${\rm{4x}} + 3{\rm{y}} - {\rm{21}} = 0$ . The new position for C is (3, 3).
So, $CB' = CA' = 5$ (length of tangents to the circles from C) We observe that $AC = 20,\,BC = 15,\,AB = 25$ . Hence the required perimeter is $50 + 5\sqrt 2$

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