Manipal Entrance Test 2020 Math 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
⇒‌‌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.....(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).