Geo4.106

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Prove that the two circles which pass through the points (0,a),(0,-a)\, and touch the line y=mx+c\, will cut orthogonally if c^{2}=a^{2}(2+m^{2})\,

Let the equation be x^{2}+y^{2}+2gx+2fy+k=0\,

Given the circle passses through the points

a^{2}+2af+k=0,a^{2}-2af+k=0\,

Solving these two

f=0,k=-a^{2}\,

Given the equation touches the line y=mx+c,mx-y+c=0\,

{\frac  {|m(-g)+c|}{{\sqrt  {m^{2}+1}}}}={\sqrt  {g^{2}-k}}\,

g^{2}+2cgm+a^{2}(m^{2}+1)-c^{2}=0\,

Let g1,g2 be the roots of the quadratic in g.

g_{1}g_{2}=a^{2}(m^{2}+1)-c^{2}\,

The equations of the two circles are x^{2}+y^{2}+2g_{1}x-a^{2}=0,x^{2}+y^{2}+2g_{2}x-a^{2}=0\,

If these two circles cut orthogonally,

2(g_{1}g_{2}+0)=-a^{2}-a^{2}\,

a^{2}(m^{2}+1)-c^{2}=-a^{2}\,

a^{2}(2+m^{2})=c^{2}\,


Main Page:Geometry:Circles