Let the tangents at the points A (4, -11) and | vedclass

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(D) The equation of the circle is $x^2 + y^2 - 3x + 10y - 15 = 0$. The equation of the tangent at $A(x_1, y_1)$ is $xx_1 + yy_1 - \frac{3}{2}(x + x_1)

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$\frac{2 \sqrt{13}}{3}$

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(D) The equation of the circle is $x^2 + y^2 - 3x + 10y - 15 = 0$. The equation of the tangent at $A(x_1, y_1)$ is $xx_1 + yy_1 - \frac{3}{2}(x + x_1) + 5(y + y_1) - 15 = 0$. For $A(4, -11)$: $4x - 11y - \frac{3}{2}(x + 4) + 5(y - 11) - 15 = 0$ $\Rightarrow 8x - 22y - 3x - 12 + 10y - 110 - 30 = 0$ $\Rightarrow 5x - 12y - 152 = 0$. For $B(8, -5)$: $8x - 5y - \frac{3}{2}(x + 8) + 5(y - 5) - 15 = 0$ $\Rightarrow 16x - 10y - 3x - 24 + 10y - 50 - 30 = 0$ $\Rightarrow 13x = 104$ $\Rightarrow x = 8$. Substituting $x = 8$ into the first tangent equation: $5(8) - 12y - 152 = 0$ $\Rightarrow 40 - 152 = 12y$ $\Rightarrow 12y = -112$ $\Rightarrow y = -\frac{28}{3}$. So,$C = (8, -\frac{28}{3})$. The line $AB$ passes through $(4, -11)$ and $(8, -5)$. The slope $m = \frac{-5 - (-11)}{8 - 4} = \frac{6}{4} = \frac{3}{2}$. The equation of line $AB$ is $y + 5 = \frac{3}{2}(x - 8)$ $\Rightarrow 2y + 10 = 3x - 24$ $\Rightarrow 3x - 2y - 34 = 0$. The radius $r$ is the perpendicular distance from $C(8, -\frac{28}{3})$ to the line $3x - 2y - 34 = 0$: $r = \frac{|3(8) - 2(-\frac{28}{3}) - 34|}{\sqrt{3^2 + (-2)^2}} = \frac{|24 + \frac{56}{3} - 34|}{\sqrt{13}} = \frac{|\frac{56}{3} - 10|}{\sqrt{13}} = \frac{|\frac{26}{3}|}{\sqrt{13}} = \frac{26}{3 \sqrt{13}} = \frac{2 \sqrt{13}}{3}$.

Let the tangents at the points $A (4, -11)$ and $B (8, -5)$ on the circle $x^2 + y^2 - 3x + 10y - 15 = 0$ intersect at the point $C$. Then the radius of the circle,whose center is $C$ and the line joining $A$ and $B$ is its tangent,is equal to

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