AP EAMCET Engineering 26 Apr 2017 Shift 1 Paper

Consider the figure shown below.Let (x1,y1) lie on the line 2kx+3y−1=0
The equation of the chord to the circle is,
x2+y2−2x−4y−4=0
And,
xx1+yy1−

2(x+x1)

2

−

4 (y+y1)

2

−4=0
x(x1−1)+y(y1−2) −x1−2y1−4=0
The lines 2x+y+5=0 and 2kx+3y+1=0 are conjugate lines.
So,
2x+y+5=0
And
x (x1−1)+y(y1−2)−x1−2y1−4=0 coincides.
Therefore,

2

x1−1

=

1

y1−2

=

−5

x1+2y1+4

=

1

λ

So,
x1=2λ+1
y1=λ+2
x1+y1+4=−5λ
On solving the above three equations.
λ=−1
x1=−1
y1=1
Thus, the value of k is obtained as,
2kx+3y−1=0
−2k+3−1=0
k=1