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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...

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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

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

3 Answers 3

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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 $(-6,-7)$.

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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., $P=(-6,-7).$

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  • $\begingroup$ 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. $\endgroup$
    – Will R
    Commented Oct 6, 2020 at 2:47
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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}

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