If a circle C passing through the point (4, 0 | vedclass

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(C) The equation of the given circle is $x^2 + y^2 + 4x - 6y - 12 = 0$. The tangent to this circle at the point $(1, -1)$ is given by $x(1) + y(-1) +

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$5$

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(C) The equation of the given circle is $x^2 + y^2 + 4x - 6y - 12 = 0$. The tangent to this circle at the point $(1, -1)$ is given by $x(1) + y(-1) + 2(x + 1) - 3(y - 1) - 12 = 0$. Simplifying this,we get $x - y + 2x + 2 - 3y + 3 - 12 = 0$,which is $3x - 4y - 7 = 0$. The family of circles touching the given circle at $(1, -1)$ is $(x - 1)^2 + (y + 1)^2 + \lambda(3x - 4y - 7) = 0$. Since the circle passes through $(4, 0)$,we substitute these coordinates: $(4 - 1)^2 + (0 + 1)^2 + \lambda(3(4) - 4(0) - 7) = 0$. This gives $9 + 1 + \lambda(12 - 7) = 0$,so $10 + 5\lambda = 0$,which implies $\lambda = -2$. Substituting $\lambda = -2$ back into the equation: $(x^2 - 2x + 1) + (y^2 + 2y + 1) - 6x + 8y + 14 = 0$. This simplifies to $x^2 + y^2 - 8x + 10y + 16 = 0$. The radius of this circle is $\sqrt{g^2 + f^2 - c} = \sqrt{(-4)^2 + (5)^2 - 16} = \sqrt{16 + 25 - 16} = \sqrt{25} = 5$.

If a circle $C$ passing through the point $(4, 0)$ touches the circle $x^2 + y^2 + 4x - 6y = 12$ externally at the point $(1, -1)$,then the radius of $C$ is

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