TS EAMCET 20-July-2022 Shift 1 Solved Paper

Line x+2y=1 cuts the X-axis at A and Y-axis at B.
Then, A(1,0) and B(0,‌

1

2

)
A circle passes through A and B and origin.
So, equation of circle is x2+y2+2gx+2fy=0
It passes through the point A(1,0),
1+0+2g=0⇒g=−‌

1

2

Also, passes through (0,‌

1

2

)2
0+‌

1

4

+0+f=0⇒f=−‌

1

4

So, equation of circle is x2+y2−x−‌

y

2

=0
So, tangent at (0,0) is
‌xx1+yy1−(‌

x+x1

2

)−‌

1

2

(‌

y+y1

2

)=0
‌0+0−(‌

x+0

2

)−‌

1

2

(‌

y+0

2

)=0
⇒‌‌2x+y=0
Perpendicular distance from A(1,0) to tangent at origin =|‌

2+0

√5

|=‌

2

√5

. Perpendicular distance from B(0,‌

1

2

) to tangent at origin
=|‌

0+ 1 2

√5

|=‌

1

2√5

Sum of perpendicular distances from A and B to tangent at origin
=(2+‌

1

2

)‌

1

√5

=‌

√5

2

Centre (−y1,−f)=(‌

1

2

,‌

1

4

)
‌‌ Radius ‌=√g2+f2−c=√‌

1

4

+‌

1

16

−0=‌

√5

4

‌‌ Diameter ‌=2r=‌

√5

2

So, sum of perpendicular distances from A and B equal to the diameter of circle.