Geo4.66

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Show that the pair of tangents drawn from (g,f)\, to the circlesx^{2}+y^{2}+2gx+2fy+c=0\, are at right angles if g^{2}+g^{2}+c=0\,

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

Equation to the pair of tangents from the point is

[S_{1}]^{2}=SS_{{11}}\,

(g^{2}+f^{2}+c+g(x+g)+f(y+f))^{2}=((x^{2}+y^{2}+2gx+2fy+c)(3g^{2}+3f^{2}+c)\,

(gx+fy+2g^{2}+2f^{2}+c)^{2}=((x^{2}+y^{2}+2gx+2fy+c)(3g^{2}+3f^{2}+c)\,

For the tangents to be at right angles, the condition is that

coefficient of x^{2}\,+ coefficient of y^{2}\,=0\,

Hence from the above

coefficient of x^{2}=g^{2}-3g^{2}-3f^{2}-c+g^{2}+f^{2}\, coefficient of y^{2}=f^{2}-3g^{2}-3f^{2}-c+2g^{2}+f^{2}\,

Hence

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

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


Main Page:Geometry:Circles