TG EAMCET 2 May 2025 Shift 1 Paper

Given circle, $x-2=5 \cos \theta, y+1=5 \sin \theta$, where $\theta$ is the perimeter.$\Rightarrow (x-2)^2+(y+1)^2=(5 \cos \theta)^2+(5 \sin \theta)^2$$=25 (\cos^2 \theta + \sin^2 \theta) =25$$\therefore$ The circle has a centre at $(h, k)=(2,-1)$ and a radius, $r_c=5$Now, the line equation is $x=1+\frac{r}{2}$ and $y=-2+\frac{\sqrt{3}}{2}r$. This is parametric form.Let $x_1=1, y_1=-2$The direction vector of the line is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$So, the slope of the line, $m=\frac{\Delta y}{\Delta x}$$\Rightarrow \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$Now, the equation of line is$y-y_1=m(x-x_1)$$\Rightarrow y-(-2)=\sqrt{3}(x-1)$$\Rightarrow y+2=\sqrt{3}x-\sqrt{3}$$\Rightarrow \sqrt{3}x-y-(2+\sqrt{3})=0$Now, calculate the distance from the center of the circle to line,$d=\frac{|A x_0+B y_0+C|}{\sqrt{A^2+B^2}}$Here, $(x_0, y_0)=(2,-1)$ and the line is $A=\sqrt{3}, B=-1, C=-(2+\sqrt{3})$So, $d=\frac{|\sqrt{3}(2)+(-1)(-1)-(2+\sqrt{3})|}{\sqrt{(\sqrt{3})^2+(-1)^2}}$$=\frac{|2\sqrt{3}+1-2-\sqrt{3}|}{\sqrt{3+1}}$$=\frac{\sqrt{3}-1}{2} \approx 0.366 < 5$$\;=\;\frac{\sqrt{3}-1}{2} \approx 0.366 < 5$So, the line intersects the circle at two distinct points, meaning it is a secant to the circle.So, the line is a chord of circle other than diameter.