If A, B are the points of contact of the tang | vedclass

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(B) The equation of the circle is $x^2+y^2-8x-10y+5=0$.Completing the square,we get $(x-4)^2+(y-5)^2 = 16+25-5 = 36 = 6^2$.Thus,the center $C$ is $(4,

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$\frac{24}{7}$

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(B) The equation of the circle is $x^2+y^2-8x-10y+5=0$.Completing the square,we get $(x-4)^2+(y-5)^2 = 16+25-5 = 36 = 6^2$.Thus,the center $C$ is $(4, 5)$ and the radius $r = 6$.The length of the tangent $PA$ from $P(-2, -3)$ is $\sqrt{S_1} = \sqrt{(-2)^2+(-3)^2-8(-2)-10(-3)+5} = \sqrt{4+9+16+30+5} = \sqrt{64} = 8$.In the right-angled triangle $	riangle CAP$,$\angle CAP = 90^\circ$.Let $\angle APC = \frac{	heta}{2}$. Then $	an(\frac{	heta}{2}) = \frac{CA}{PA} = \frac{6}{8} = \frac{3}{4}$.Using the formula $	an 	heta = \frac{2 	an(\frac{	heta}{2})}{1-	an^2(\frac{	heta}{2})}$,we have:$	an 	heta = \frac{2(\frac{3}{4})}{1-(\frac{3}{4})^2} = \frac{\frac{3}{2}}{1-\frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{3}{2} 	imes \frac{16}{7} = \frac{24}{7}$.

If $A, B$ are the points of contact of the tangents drawn from the point $P(-2, -3)$ to the circle $x^2+y^2-8x-10y+5=0$ and the chord $AB$ subtends an angle $\theta$ at $P$,then $\tan \theta =$

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