The equation of the circle having the lines x | vedclass

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(B) The given circle is $x(x - 4) + y(y - 3) = 0$,which simplifies to $x^2 + y^2 - 4x - 3y = 0$.Its center $C_1$ is $(2, 3/2)$ and radius $r_1 = \sqrt

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$x^2 + y^2 + 6x - 3y - 45 = 0$

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(B) The given circle is $x(x - 4) + y(y - 3) = 0$,which simplifies to $x^2 + y^2 - 4x - 3y = 0$.Its center $C_1$ is $(2, 3/2)$ and radius $r_1 = \sqrt{2^2 + (3/2)^2} = \sqrt{4 + 9/4} = \sqrt{25/4} = 5/2$.The normals are given by $x^2 + 2xy + 3x + 6y = 0$,which factors as $x(x + 2y) + 3(x + 2y) = 0$,or $(x + 3)(x + 2y) = 0$.The intersection of these lines gives the center $C_2$ of the required circle. Solving $x + 3 = 0$ and $x + 2y = 0$,we get $x = -3$ and $y = 3/2$. So $C_2 = (-3, 3/2)$.The distance between centers $C_1(2, 3/2)$ and $C_2(-3, 3/2)$ is $d = \sqrt{(2 - (-3))^2 + (3/2 - 3/2)^2} = 5$.Since the circle $C_2$ contains $C_1$,the condition is $d = r_2 - r_1$,where $r_2$ is the radius of the required circle.$5 = r_2 - 5/2 \Rightarrow r_2 = 5 + 5/2 = 15/2$.The equation of the circle is $(x + 3)^2 + (y - 3/2)^2 = (15/2)^2$.$x^2 + 6x + 9 + y^2 - 3y + 9/4 = 225/4$.$x^2 + y^2 + 6x - 3y + 9 + 9/4 - 225/4 = 0$.$x^2 + y^2 + 6x - 3y + 9 - 216/4 = 0$.$x^2 + y^2 + 6x - 3y + 9 - 54 = 0$.$x^2 + y^2 + 6x - 3y - 45 = 0$.

The equation of the circle having the lines $x^2 + 2xy + 3x + 6y = 0$ as its normals and having a size just sufficient to contain the circle $x(x - 4) + y(y - 3) = 0$ is

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