For the circle x-2=5 cos theta, y+1=5 sin the | vedclass

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(A) The given circle is $(x-2)^2 + (y+1)^2 = 25$,which has center $C(2, -1)$ and radius $R = 5$. The line is given by $x = 1 + \frac{r}{2}$ and $y = -

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Chord of the circle other than diameter

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(A) The given circle is $(x-2)^2 + (y+1)^2 = 25$,which has center $C(2, -1)$ and radius $R = 5$. The line is given by $x = 1 + \frac{r}{2}$ and $y = -2 + \frac{\sqrt{3}}{2} r$. Rearranging for $r$,we get $r = 2(x-1)$ and $r = \frac{2}{\sqrt{3}}(y+2)$. Equating these,$2(x-1) = \frac{2}{\sqrt{3}}(y+2) \implies \sqrt{3}x - y - (\sqrt{3}+2) = 0$. The perpendicular distance $d$ from the center $(2, -1)$ to the line is $d = \frac{|\sqrt{3}(2) - (-1) - (\sqrt{3}+2)|}{\sqrt{(\sqrt{3})^2 + (-1)^2}} = \frac{|2\sqrt{3} + 1 - \sqrt{3} - 2|}{\sqrt{3+1}} = \frac{|\sqrt{3}-1|}{2}$. Since $d Since the line does not pass through the center $(2, -1)$ (substituting $x=2, y=-1$ gives $\sqrt{3}(2) - (-1) - \sqrt{3} - 2 = \sqrt{3}-1 
eq 0$),it is a chord other than the diameter.

For the circle $x-2=5 \cos \theta, y+1=5 \sin \theta$ where $\theta$ is the parameter,the line $x=1+\frac{r}{2}, y=-2+\frac{\sqrt{3}}{2} r$ where $r$ is the parameter,is a

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