Geo4.105

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Find the equation of the circle which cut orthogonally the circles x^{2}+y^{2}-6y+1=0,x^{2}+y^{2}-4y+1=0\, and touch the line 3x+4y+5=0\,

Let the required circle bex^{2}+y^{2}+2gx+2fy+c=0\,

This circle cuts the two circles orthogonally,hence

2g(0)+2f(-3)=c+1,2g(0)+2f(-2)=c+1\,

-6f=c+1,-4f=c+1\,

Solving these equations we get

f=0,c=-1\,

Given the required circle also touches the line 3x+4y+5=0\,

Now

{\frac  {|-3g-4f+5|}{{\sqrt  {9+16}}}}={\sqrt  {g^{2}+f^{2}-c}}\,

{\frac  {|-3g+5|}{5}}={\sqrt  {g^{2}+1}}\,

16g^{2}+30g=0,g=0,{\frac  {-15}{8}}\,

Therefore the required circle equationis

x^{2}+y^{2}-1=0,4(x^{2}+y^{2})-15x-4=0\,

Main Page:Geometry:Circles