Geo4.70

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Find the condition that the pair of tangents from the origin to the circle x^{2}+y^{2}+2gx+2fy+c=0\, may be at right angles.

Given Circle is x^{2}+y^{2}+2gx+2fy+c=0\,

Now the equation of the pair of tangents from the origin (0,0) is

(gx+fy+c)^{2}=(x^{2}+y^{2}+2gx+2fy+c)c\,

g^{2}x^{2}+f^{2}y^{2}+c^{2}+2gfxy+2fcy+2gcx=cx^{2}+cy^{2}+2gcx+2fcy+c^{2}\,

Simplifying, we have

(g^{2}-c)x^{2}+2fgxy+(f^{2}-c)y^{2}=0\,

Sum of  the Coefficient of x^{2}\, and y^{2}\, is

(g^{2}-c+f^{2}-c)=0\,

g^{2}+f^{2}=2c\,


Main Page:Geometry:Circles