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

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

What I tried...

The equation $$x^2=y-6$$ is of a parabola. To find the slope of the tangent to the parabola at the point $$(1,7)$$, $$\frac{dy}{dx}\Bigg|_{(1,7)}=2\tag{Slope of the line tangent to the parabola}$$ So the equation of the line is $$2x-y+5=0\implies y=2x+5$$

Substituting this in the equation of circle to find the point of intersection of the line with the circle, we get, $$x^2+(2x+5)^2+16x+12(2x+5)+c=0$$ Solving this, I get a complicated equation and then the answer comes out in terms of $$c$$ but the actual answer does not contain $$c$$ at all.
I would prefer a more analytical/geometrical approach if possible

• Did you write the whole problem? I think the problem statement is incomplete. Can you clarify your question?
– user798113
Commented Oct 6, 2020 at 2:07
• @Ramanujan That was the whole problem
– rash
Commented Oct 6, 2020 at 2:08
• +1 Interesting problem, very nicely presented, outstanding work shown. Better than 95% of the queries posted at mathSE. Commented Oct 6, 2020 at 2:15

Because the line "touches" the circle, there is only one point of intersection with the circle. Therefore, the equation, $$x^2+(2x+5)^2+16x+12(2x+5)+c=0 \implies 5 x^2 + 60 x + 85 + c= 0$$ has one solution.

So the determinant $$\Delta = 3600 - 4\cdot5\cdot(85+c)=0$$. Ergo, $$c=95$$.

And solving the equation $$5x^2 + 60x +180=0$$ gives the solution $$(-6,-7)$$.

By rearranging the equation of the circle, we quickly find that the centre of the circle is $$(-8,-6)$$ (as you have illustrated).

Let $$O$$ be the centre of the circle, and $$P$$ be the point where the line $$y=2x+5$$ and circle touch. Since the line and circle touch at $$P,$$ the line through $$O$$ and $$P$$ must be the line perpendicular to $$y=2x+5$$ at $$P$$ (that is, the radius is perpendicular to the tangent).

So now we're looking for a line perpendicular to $$y=2x+5$$ and passing through $$O=(-8,-6).$$ By vector geometry, this line has parametric form $$r(t) = (-8,-6) + t(-2,1),$$ i.e., $$r(t) = (-2t-8,t-6).$$ Now set $$x=-2t-8$$ and $$y=t-6$$ in $$y=2x+5$$ and solve for $$t$$: we find $$t=1,$$ i.e., $$P=(-6,-7).$$

• The insight behind this solution is that even if you forget the circle, then the point $P$ still remains: it's just the point obtained by dropping a perpendicular from $O$ to the tangent line. From this point of view, the circle, and hence its radius, are actually irrelevant. Commented Oct 6, 2020 at 2:47

As you've simplified the question, we need to find a point, where the line \begin{align} y&=2x+5 \tag{1}\label{1} \end{align}
touches the circle of unknown radius centered at $$O=(-8,-6)$$.

A convenient point on the tangent line below the center $$O$$ is $$A(-8,-11)$$, $$|OA|=5$$.

Let $$\phi=\angle T_cOA$$,

\begin{align} \phi&=\arctan2 =\arccos\tfrac{\sqrt5}5 =\arcsin\tfrac{2\sqrt5}5 \tag{2}\label{2} . \end{align}

Then the radius of the circle

\begin{align} r=|OT_c|&=|OA|\cdot\cos\phi =\sqrt5 \tag{3}\label{3} ,\\ T_c&=O+r\cdot(\sin\phi,\,-\cos\phi) \\ &=O+(2,-1)=(-6,-7) . \end{align}