Tangent to the curve y=x2+6 at a point P (1,7) touches the circle x 2+ y 2+16 x +12 y + c =0 at a point Q. Then the coordinates of Q are

Question

Tangent to the curve y=x2+6 at a point P (1,7) touches the circle x 2+ y 2+16 x +12 y + c =0 at a point Q. Then the coordinates of Q are (A) (-6,-11) (B) (-9,-13) (C) (-10,-15) (D) (-6,-7)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/94f818a82085afa6f185092c178e95b4-.png" /> Equation of tangent to the parabola at (1,7) is x-((y+7)/2)+6=0 ⇒ 2 x-y+5=0 ⇒ Centre =(-8,-6) Equation of C Q-x+2 y+k=0 =8-12+k=0 ⇒ k=20 P Q=4 x-2 y+10=0 C Q=x+2 y+20=0 =5 x+30=0 ⇒ x=-6 ⇒ -6+2 y+20-0 ⇒ y=-7 Hence the point of coatact is (-6,-7).

Q. Tangent to the curve $y = x^{2} + 6$ at a point $P (1, 7)$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at a point $Q$ . Then the coordinates of $Q$ are

$(- 6, - 11)$

$(- 9, - 13)$

$(- 10, - 15)$

$(- 6, - 7)$

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

(−6,−11)

(−9,−13)

(−10,−15)

(−6,−7)

Equation of tangent to the parabola at (1,7) is x−2(y+7)​+6=0⇒2x−y+5=0 ⇒ Centre=(−8,−6) Equation of CQ−x+2y+k=0 =8−12+k=0⇒k=20 PQ=4x−2y+10=0 CQ=x+2y+20=0 =5x+30=0⇒x=−6 ⇒−6+2y+20−0⇒y=−7 Hence the point of coatact is (−6,−7).

Q. Tangent to the curve $y = x^{2} + 6$ at a point $P (1, 7)$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at a point $Q$ . Then the coordinates of $Q$ are

$(- 6, - 11)$

$(- 9, - 13)$

$(- 10, - 15)$

$(- 6, - 7)$