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If the angle between the pair of tangents drawn to the circle $x^2+y^2-2 x+4 y+3=0$ from the point $(6,-5)$ is $\theta$, then $\cot \theta=$
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The correct answer is:
$\frac{15}{8}$
$x^2+y^2-2 x+4 y+3=0$
$\mathrm{PA}=$ length of tangent
$\begin{aligned} & =\sqrt{\mathrm{S}_1} \\ & =\sqrt{6^2+(-5)^2-2(6)+4(-5)+3}=\sqrt{32}=4 \sqrt{2} \\ & r=\sqrt{1^2+(-2)^2-3}=\sqrt{2} \\ & \cot \frac{\theta}{2}=\frac{\mathrm{L}}{r}=\frac{4 \sqrt{2}}{\sqrt{2}}=4 \\ & \tan \frac{\theta}{2}=\frac{1}{4} \\ & \tan \theta=\frac{2 \tan \frac{\theta}{2}}{1-\tan ^2 \frac{\theta}{2}}=\frac{2\left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}\end{aligned}$
$=\frac{1}{2} \times \frac{16}{15}=\frac{8}{15} \quad \therefore \cot \theta=\frac{15}{8}$
$\mathrm{PA}=$ length of tangent
$\begin{aligned} & =\sqrt{\mathrm{S}_1} \\ & =\sqrt{6^2+(-5)^2-2(6)+4(-5)+3}=\sqrt{32}=4 \sqrt{2} \\ & r=\sqrt{1^2+(-2)^2-3}=\sqrt{2} \\ & \cot \frac{\theta}{2}=\frac{\mathrm{L}}{r}=\frac{4 \sqrt{2}}{\sqrt{2}}=4 \\ & \tan \frac{\theta}{2}=\frac{1}{4} \\ & \tan \theta=\frac{2 \tan \frac{\theta}{2}}{1-\tan ^2 \frac{\theta}{2}}=\frac{2\left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}\end{aligned}$
$=\frac{1}{2} \times \frac{16}{15}=\frac{8}{15} \quad \therefore \cot \theta=\frac{15}{8}$
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