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(C) The equation of the normals is given by $x^2-3xy-3x+9y=0$. Factoring this,we get $x(x-3y)-3(x-3y)=0$,which implies $(x-3y)(x-3)=0$. Thus,the two n

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$x^2+y^2-6x-2y+1=0$

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(C) The equation of the normals is given by $x^2-3xy-3x+9y=0$. Factoring this,we get $x(x-3y)-3(x-3y)=0$,which implies $(x-3y)(x-3)=0$. Thus,the two normals are $x-3y=0$ and $x=3$. The centre of the circle is the intersection of these two normals: $x=3$ and $x=3y$,which gives $y=1$. So,the centre of the required circle is $(3,1)$. The given circle is $x^2+y^2-6x+6y+17=0$. Its centre is $(3,-3)$ and its radius $r_1 = \sqrt{3^2+(-3)^2-17} = \sqrt{9+9-17} = \sqrt{1} = 1$. Since the circles touch externally,the distance between the centres equals the sum of the radii: $r_1+r_2 = \sqrt{(3-3)^2+(1-(-3))^2} = \sqrt{0^2+4^2} = 4$. Substituting $r_1=1$,we get $1+r_2=4$,so $r_2=3$. The equation of the circle with centre $(3,1)$ and radius $3$ is $(x-3)^2+(y-1)^2=3^2$. Expanding this,$x^2-6x+9+y^2-2y+1=9$,which simplifies to $x^2+y^2-6x-2y+1=0$.

The equation of the circle which touches the circle $x^2+y^2-6x+6y+17=0$ externally and has the lines $x^2-3xy-3x+9y=0$ as two normals,is

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