The tangent at (1, 7) to the curve x2 = y − 6 touches the circle x2 +y2 + 16x + 12y + c = 0 at

Question

The tangent at (1, 7) to the curve x2 = y − 6 touches the circle x2 +y2 + 16x + 12y + c = 0 at (A) (6, 7) (B) (− 6, 7) (C) (6, − 7) (D) (− 6, − 7)

Tardigrade Admin · Accepted Answer

The tangent at (1,7) to the parabola x2=y-6 is x =(1/2)(y+7)-6 ⇒ 2 x =y+7-12 ⇒ y =2 x+5 which is also a tangent to the circle x2+y2+16 x+12 y +c=0 ∴ x2+(2 x+5)2+ 16 x+12(2 x+5)+c=0 ⇒ 5 x2+60 x+85+c=0 must have equal roots. Let α and β be the roots of the equation. Then, α+β=-12 ⇒ α=-6 (∵ α=β) ∴ x=-6 and y=2 x+5=-7 ⇒ Point of contact is (-6,-7)

Q. The tangent at $(1, 7)$ to the curve $x^{2} = y - 6$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at

$(6, 7)$

$(- 6, 7)$

$(6, - 7)$

$(- 6, - 7)$

Q. The tangent at (1,7) to the curve x2=y−6 touches the circle x2+y2+16x+12y+c=0 at

(6,7)

(−6,7)

(6,−7)

(−6,−7)

Q. The tangent at $(1, 7)$ to the curve $x^{2} = y - 6$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at

$(6, 7)$

$(- 6, 7)$

$(6, - 7)$

$(- 6, - 7)$