Question

Tangent to the curve y = x² +6 at a point (1, 7) touches the circle x² +y² +16x+12y+c = 0 at a point Q then coordinates of Q are

a. (–6, –11)

b. (–9, –13)

c. (–10. –15)

d. (–6, –7)

Answer 1

Correct option is d. (–6, –7)

Equation of tangent at (1, 7) to the curve y = x² +6 is

\(\frac{y+7}2=x+6\)

2x – y+5 = 0 .....(1)

This line also touches the circle i.e.

Equation of normal of circle passing through centre (–8, –6).

x+2y + λ = 0

–8 –12 + λ = 0

λ = 20

∴x + 2y + 20 = 0 .....(2)

Q is intersection point of (1) and (2)

x = –6, y = –7

Q(–6, –7)