If the angle between the pair of tangents dra | vedclass

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(C) The equation of the circle is $x^2+y^2-2x+4y+3=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=2, c=3$. The center of the circle is $O(-g

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$\frac{8}{15}$

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(C) The equation of the circle is $x^2+y^2-2x+4y+3=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=2, c=3$. The center of the circle is $O(-g, -f) = (1, -2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+(2)^2-3} = \sqrt{1+4-3} = \sqrt{2}$. Let $P$ be the point $(6, -5)$. The length of the tangent $PA$ is given by $\sqrt{S_1}$,where $S_1 = x_1^2+y_1^2-2x_1+4y_1+3$. $PA = \sqrt{(6)^2+(-5)^2-2(6)+4(-5)+3} = \sqrt{36+25-12-20+3} = \sqrt{32} = 4\sqrt{2}$. In the right-angled triangle $	riangle OAP$,where $\angle OAP = 90^\circ$,we have $	an(	heta/2) = \frac{OA}{AP} = \frac{r}{PA} = \frac{\sqrt{2}}{4\sqrt{2}} = \frac{1}{4}$. Using the formula $	an 	heta = \frac{2	an(	heta/2)}{1-	an^2(	heta/2)}$,we get: $	an 	heta = \frac{2(1/4)}{1-(1/4)^2} = \frac{1/2}{1-1/16} = \frac{1/2}{15/16} = \frac{1}{2} 	imes \frac{16}{15} = \frac{8}{15}$.

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

If the angle between the pair of tangents drawn to the circle $x^2+y^2-2x+4y+3=0$ from $(6,-5)$ is $\theta$,then $\tan \theta=$

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