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=−