Geo4.13

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One end of thediameter of the circlex^{2}+y^{2}-10x-2y+6=0\, is (3,5).Find the other end of the diameter.

Centre of the circle is

\left({\frac  {-(-10)}{2}},{\frac  {-(-2)}{2}}\right)\,

\left(5,1\right)\,

Centre is the mid point of the diameter,let the other end be (x,y)

Now

\left({\frac  {x+3}{2}},{\frac  {y+5}{2}}\right)=(5,1)\,

Comparing the coordinates, we get

{\frac  {x+3}{2}}=5,{\frac  {y+5}{2}}=1\,

x+3=10,y+5=2\,

x=-7,y=-3\,

Hence the other end of the diameter is (-7,-3)


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