The circle x^2 + y^2 - 8x = 0 and the hyperbo | vedclass

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(B) The circle is $x^2 + y^2 - 8x = 0$,which can be written as $(x-4)^2 + y^2 = 16$. Its center is $(4, 0)$ and radius $r = 4$. Let the tangent to the

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$2x - \sqrt{5}y + 4 = 0$

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(B) The circle is $x^2 + y^2 - 8x = 0$,which can be written as $(x-4)^2 + y^2 = 16$. Its center is $(4, 0)$ and radius $r = 4$. Let the tangent to the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ be at point $P(3 \sec 	heta, 2 	an 	heta)$. The equation of the tangent is $\frac{x \sec 	heta}{3} - \frac{y 	an 	heta}{2} = 1$. Since this is a tangent to the circle,the perpendicular distance from the center $(4, 0)$ to this line must be equal to the radius $r = 4$. $\frac{|\frac{4 \sec 	heta}{3} - 1|}{\sqrt{\frac{\sec^2 	heta}{9} + \frac{	an^2 	heta}{4}}} = 4$. Squaring both sides: $\frac{(\frac{4}{3} \sec 	heta - 1)^2}{\frac{4 \sec^2 	heta + 9 	an^2 	heta}{36}} = 16$. $\frac{16}{9} \sec^2 	heta + 1 - \frac{8}{3} \sec 	heta = 16 \cdot \frac{4 \sec^2 	heta + 9(\sec^2 	heta - 1)}{36} = \frac{4}{9} (13 \sec^2 	heta - 9)$. $16 \sec^2 	heta + 9 - 24 \sec 	heta = 52 \sec^2 	heta - 36$. $36 \sec^2 	heta + 24 \sec 	heta - 45 = 0$. Dividing by $3$: $12 \sec^2 	heta + 8 \sec 	heta - 15 = 0$. $(6 \sec 	heta - 5)(2 \sec 	heta + 3) = 0$. Since $\sec 	heta = 5/6$ is impossible,we have $\sec 	heta = -3/2$. Then $	an^2 	heta = \sec^2 	heta - 1 = \frac{9}{4} - 1 = \frac{5}{4}$,so $	an 	heta = \pm \frac{\sqrt{5}}{2}$. For a positive slope,the tangent equation $\frac{x \sec 	heta}{3} - \frac{y 	an 	heta}{2} = 1$ requires the coefficient of $y$ to be negative,so $	an 	heta = -\frac{\sqrt{5}}{2}$. Substituting $\sec 	heta = -3/2$ and $	an 	heta = -\sqrt{5}/2$: $\frac{x(-3/2)}{3} - \frac{y(-\sqrt{5}/2)}{2} = 1$ $\Rightarrow -\frac{x}{2} + \frac{\sqrt{5}y}{4} = 1$ $\Rightarrow -2x + \sqrt{5}y = 4$ $\Rightarrow 2x - \sqrt{5}y + 4 = 0$.

The circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ intersect at points $A$ and $B$. The equation of the common tangent with a positive slope to the circle and the hyperbola is:

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