Question

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

Options

• (0, 7)

• (-6, 7)

• (6, -7)

• (-6, -7)

Answer 1

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

Explanation:

The tangent at (1, 7) to the parabola 

\[x^{2}=y-6x\mathrm{is}\] is

                    \[x(1)=\frac{1}{2}\left(y+7\right)-6\]

\[\{\text{Replacing }x^2\to xx_1,2y\to y+y_1\}\]

\[\Rightarrow\quad y=2x+5\]

which is also tangent to the circle.

\[x^2+y^2+16x+12y+c=0\]

\[\therefore x^2+(2x+5)^2+16x+12(2x+5)+c=0\]

have equal and real roots

\[\Rightarrow\quad5x^2+60x+85+c=0\]

\[\Rightarrow\alpha=\beta\]

\[\Rightarrow\quad\alpha+\beta=\frac{-b}{a}\Rightarrow2\alpha=\frac{-60}{5}\]

\[\Rightarrow\quad\alpha=-6\]

\[\therefore\quad x=-6\]

\[andy=2x+5=2(-6)+5\]

                      \[y=-7\]

\[\therefore\] Point of contact (-6, -7)