Tangent line circle

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Find the circle equation which passes through point(4,1) and one of the tangent line to this circle is x+3y=3 Solution : Circle equation which its center(4,1) and its radius=0 (x-4)^2+(y-1)^2=0 x^2+y^2-8x-2y+16+1=0 x^2+y^2-8x-2y+17=0 p[x+3y-3]=0 -------------------- + x^2+y^2+(p-8)x+(3p-2)y+(17-3p)=0 x+3y-3=0 is its tangent line this means : x1*x+y1*y+(p/2-4)*x1+(p/2-4)*x+(3p/2-1)*y1+(3p/2-1)*y+(17-3p)=0 where(x1,y1) lies on x+3y-3=0 or x1+3y1-3=0 x1=3-3y1 (3-3y1)*x++y1*y+(p/2-4)(3-3y1)+(p/2-4)*y

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