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If the point (1+\cos \theta, \sin \theta) lies between the region corresponding to the acute angle between the lines 3 y=x and 6 y=x, then
 

Option: 1

\theta \in R


Option: 2

\theta \in R-n \pi, n \in I


Option: 3

\theta \in R-(2 n+1) \frac{\pi}{2}, n \in I


Option: 4

 None of these


Answers (1)

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Let the centre of the circle be \mathrm{\left ( a,3-a \right ).}Since the circle is tangent to the line \mathrm{2x+y=5,} the distance between the centre and this line must be equal to the radius of the circle. The distance between a point (x,y) and the line \mathrm{2x+y=5,} is:

\left|\frac{2 x+y-5}{\sqrt{2^2+1^2}}\right|

Therefore, the equation we need to solve is: \left|\frac{2 a+(3-a)-5}{\sqrt{2^2+1^2}}\right|=\sqrt{(a-1)^2+(3-a-2)^2}

Simplifying this equation, we get: \frac{|3 a-8|}{\sqrt{5}}=\sqrt{a^2-4 a+5}

Squaring both sides of the equation, we get:

\begin{aligned} & 9 a^2-48 a+64=5 a^2-20 a+25 \\ & 4 a^2-28 a+39=0 \end{aligned}

Solving for we get: 

\begin{aligned} & a=\frac{28 \pm \sqrt{\left(28^2-4(4)(39)\right)}}{2(4)} \\ & a=\frac{7}{2} \text { or } \frac{13}{2} \end{aligned}

Substituting into the equation \mathrm{x+y=3,}  we get the corresponding values of \mathrm{b:}

\mathrm{b=\frac{1}{2} \text { or } \frac{5}{2}}

Therefore, the centres of the circles are \left(\frac{7}{2}, \frac{5}{2}\right) \text { and }\left(\frac{13}{2}, \frac{-1}{2}\right)

The circle that passes through (1,2) and has a centre \left(\frac{7}{2}, \frac{5}{2}\right) is not tangent to the line, 2 x+y=5. Therefore, the circle we are looking for must have a centre \left ( \frac{-13}{2} ,\frac{-1}{2}\right ) and radius \sqrt{\left(\frac{13}{2-1}\right)^2+\left(\frac{3}{2-2}\right)^2}=\sqrt{\frac{65}{2}}

The equation of the circle is:

\left(x-\frac{13}{2}\right)^2+\left(y+\frac{1}{2}\right)^2=\frac{65}{4}

Simplifying, we get:

(x-1)^2+(y-2)^2=\frac{5}{2}

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