The equation of the circle which touches the | vedclass

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(D) The given lines are ${x^2 - 3xy - 3x + 9y = 0}$,which can be factored as ${x(x - 3y) - 3(x - 3y) = 0}$,giving ${(x - 3)(x - 3y) = 0}$.Thus,the nor

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

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(D) The given lines are ${x^2 - 3xy - 3x + 9y = 0}$,which can be factored as ${x(x - 3y) - 3(x - 3y) = 0}$,giving ${(x - 3)(x - 3y) = 0}$.Thus,the normals are ${x = 3}$ and ${x = 3y}$.The intersection of these lines gives the center of the required circle,which is ${(3, 1)}$.The given circle is ${x^2 + y^2 - 6x + 6y + 17 = 0}$,which can be written as ${(x - 3)^2 + (y + 3)^2 = 1}$.Its center is ${C_1 = (3, -3)}$ and radius ${r_1 = 1}$.Let the required circle have center ${C_2 = (3, 1)}$ and radius ${r_2}$.The distance between the centers is ${C_1C_2 = \sqrt{(3 - 3)^2 + (1 - (-3))^2} = \sqrt{0^2 + 4^2} = 4}$.Since the circles touch externally,${C_1C_2 = r_1 + r_2}$.${4 = 1 + r_2 \implies r_2 = 3}$.The equation of the circle is ${(x - 3)^2 + (y - 1)^2 = 3^2}$.${x^2 - 6x + 9 + y^2 - 2y + 1 = 9}$.${x^2 + y^2 - 6x - 2y + 1 = 0}$.

The equation of the circle which touches the circle ${x^2 + y^2 - 6x + 6y + 17 = 0}$ externally and to which the lines ${x^2 - 3xy - 3x + 9y = 0}$ are normals,is

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