TG EAMCET 2 May 2025 Shift 1 Paper

Given circle, x−2=5‌cos‌θ,y+1=5sin‌θ, where θ is the perimeter.
‌⇒(x−2)2+(y+1)2=(5‌cos‌θ)2+(5sin‌θ)2
‌=25(cos2θ+sin‌2θ)=25
∴ The circle has a centre at (h,k)=(2,−1) and a radius, rc=5
Now, the line equation is x=1+‌

r

2

and y=−2+‌

√3

2

r. This is parametric form.
Let x1=1,y1=−2
The direction vector of the line is (‌

1

2

,‌

√3

2

)
So, the slope of the line, m=‌

∆y

∆x

⇒‌‌‌

‌ √3 2

1 2

=√3
Now, the equation of line is
‌y−y1=m(x−x1)
⇒y−(−2)=√3(x−1)
⇒y+2=√3x−√3
⇒√3x−y−(2+√3)=0
Now, calculate the distance from the center of the circle to line,
d=‌

|Ax0+By0+C|

√A2+B2

Here, (x0,y0)=(2,−1) and the line is √3x−y−(2+√3)=0
So, A=√3,B=−1,C=−(2+√3)
d‌=‌

|√3(2)+(−1)(−1)−(2+√3)|

√(√3)2+(−1)2

‌=‌

|2√3+1−2−√3|

√3+1

‌=‌

√3−1

2

≈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.