AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

The equation of the line is given as ax+by=1 and the
equation of the curve is given as, x2+y2−x−y−1=0

By homogenising both equations,
x2+y2−x(ax+by)−y(ax+by)−(ax+by)2=0

x2+y2−ax2−bxy−axy−by2−(a2x2+b2y2+2abxy)=0

x2(1−a−a2)+xy(−a−b−2ab)+y2(1−b−b2)=0

since it is given that the pair of lines joining the origin and the points of intersection of the line and curve are at right angles.

Coefficient of x2+ coefficient of y2=0

1−a−a2+1−b−b2=0

a2+b2+a+b−2=0

The locus of the circle from the above equation (a,b)

Now,
x2+y2+x+y−2=0

On comparing in equation

(i) with x2+y2+2gx+2fy+c=0

The radius is,

r‌‌=√g2+f2−c

‌‌=√(‌

1

2

)2+(‌

1

2

)2+2

‌‌=√‌

1

4

+‌

1

2

+2

‌‌=√‌

5

2