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(B) The given circle is $x^2+y^2-6x+6y+17=0$. Its center is $C_1 = (3, -3)$ and its radius is $r_1 = \sqrt{3^2+(-3)^2-17} = \sqrt{9+9-17} = 1$.The lin

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

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(B) The given circle is $x^2+y^2-6x+6y+17=0$. Its center is $C_1 = (3, -3)$ and its radius is $r_1 = \sqrt{3^2+(-3)^2-17} = \sqrt{9+9-17} = 1$.The lines $x^2-3xy-3x+9y=0$ are normal to the required circle. Factoring the equation: $x(x-3y) - 3(x-3y) = 0 \Rightarrow (x-3)(x-3y) = 0$. Thus,the lines are $x=3$ and $y=x/3$. The intersection of these normals is the center of the required circle,$C_2 = (3, 1)$.Since the circles touch externally,the distance between centers $d = C_1C_2 = \sqrt{(3-3)^2 + (1-(-3))^2} = \sqrt{0^2 + 4^2} = 4$. For external contact,$d = r_1 + r_2$. So,$4 = 1 + r_2$,which gives $r_2 = 3$.The equation of the required circle with center $(3, 1)$ and radius $3$ is $(x-3)^2 + (y-1)^2 = 3^2$.Expanding this: $x^2 - 6x + 9 + y^2 - 2y + 1 = 9 \Rightarrow x^2 + y^2 - 6x - 2y + 1 = 0$.

The equation of the circle touching the circle $x^2+y^2-6x+6y+17=0$ externally and to which the lines $x^2-3xy-3x+9y=0$ are normal is

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