CGPET 2015 Solved Paper

Let required point be (α,β) on the straight line y=2x+11, which is nearest to the circle
16(x2+y2)+32x−8y−50=0
⇒ x2+y2+2x−

1

2

y−

50

16

=0
∴ centre of circle =(−1,

1

4

)
and radius, r=√1+

1

16

+

50

16

=

√67

4

Now, equation of straight line passing through centre (−1,

1

4

) and (α,β) is
y−

1

4

=(

β− 1 4

α+1

)(x+1) ...(i)
Now, gradient of this straight line =(

β− 1 4

α+1

)
Since, straight line (i) is perpendicular to the line
y=2x+11
∴ (

β− 1 4

α+1

)×2=−1[∵m1.m2=−1]
⇒ 2β−

1

2

=−α−1
⇒ 2β+α=−1+

1

2

=−

1

2

⇒ 4β+2α=−1 ...(ii)
∵ Point (α,β) lies on straight line
y=2x+11
∴ β=2α+11
⇒ β−2α=11..(iii)
On solving Eqs. (ii) and (iii), we get
5β=10⇒β=2
and 2α=2−11=−9⇒α=−

9

2

∴ Required point is (−

9

2

,2)