If the angle between the pair of tangents dra | vedclass

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(D) 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 radius $r = \sqrt{g^2+f^2-c}

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$\frac{15}{8}$

Vedclass · Answer

(D) 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 radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+2^2-3} = \sqrt{1+4-3} = \sqrt{2}$. The length of the tangent $L$ from point $P(6,-5)$ to the circle is given by $\sqrt{S_1}$. $L = \sqrt{6^2+(-5)^2-2(6)+4(-5)+3} = \sqrt{36+25-12-20+3} = \sqrt{32} = 4\sqrt{2}$. In the right-angled triangle $	riangle OAP$,where $O$ is the center and $A$ is the point of contact,$	an(\frac{	heta}{2}) = \frac{r}{L} = \frac{\sqrt{2}}{4\sqrt{2}} = \frac{1}{4}$. We know that $	an 	heta = \frac{2	an(\frac{	heta}{2})}{1-	an^2(\frac{	heta}{2})} = \frac{2(\frac{1}{4})}{1-(\frac{1}{4})^2} = \frac{1/2}{1-1/16} = \frac{1/2}{15/16} = \frac{1}{2} 	imes \frac{16}{15} = \frac{8}{15}$. Therefore,$\cot 	heta = \frac{1}{	an 	heta} = \frac{15}{8}$.

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

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