AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper

Let point P(x1,y1) then

10

√x12+y12−12x1−16y1

=

1

√3

From figure,
x12+y12−12x1−16y1=300
(x1−6)2+(y1−8)2=400 …… (I)
As, line AB in a chord of contact of the circle with respect topoint P(x1,y1), so
xx1+yy1−6(x+x1)−8(y+y1)=0
x(x1−6)+y(y1−8)−(6x1+8y1)=0

x1−6

−5

=

y1−8

5

=

6x1+8y1

K

=λ
So,
x1=6−5λ, and y1=8+5λ …… (II)
Form equation (I) and (II), we get
25λ2+25λ2=400
λ=±2√2
Also, λ=

6x1+8y1

k

So,
K=

6x1+8y1

λ

=

36−30λ+64+40λ

λ

=

100

λ

+10
=10±25√2
=5(2±5√2)