Geo4.50

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Find the value of k if the points (4,2) and (8,k) are conjugate with respect to 3x^{2}+3y^{2}-12x+4y-4=0\,

Given circle is 3x^{2}+3y^{2}-12x+4y-4=0\,

x^{2}+y^{2}-4x+{\frac  {4}{3}}y-{\frac  {4}{3}}=0\,

Now the polar of (4,2) with respect to the given circle is

4x+2y-2(x+4)+{\frac  {2}{3}}(y+2)-{\frac  {4}{3}}=0\,

6x+6y-24+2y+4-4=0\,

6x+8y-24=0\,

3x+4y-8=0\,

Since the two points are conjugate,the point (8,k)lies on this line.

Hence

24+4k-8=0,4k=-16,k=-4\,


Main Page:Geometry:Circles