# Math Help - Finding the common tangents to a circle and a parabola...

1. ## Finding the common tangents to a circle and a parabola...

EDIT: After finishing writing this post, it seems to be a bit on the lengthier side, so if you don't have the time to read the whole message, I'm just asking for some help in solving the following system of two equations in two unknowns:

$-2c=\frac{d}{\sqrt{4-d^2}}$

$-c^2-4-c(-2c)=-\sqrt{4-d^2}-d\frac{d}{\sqrt{4-d^2}}$.

But if you're curious how I arrived at this exciting system, read on...

Hi,

I'd be really grateful for some help with this problem:

Find the common tangents to the following curves:
$y+x^2=-4$
$x^2+y^2=4$

When drawn, these curves would look like this:

and our task is to find a line which is tangent to both the circle and the parabola. There will, of course, be two such lines, but that should follow from the calculation.

Let's call the point where the tangent touches the parabola ' $C$', and
the point where the tangent touches the circle ' $D$'.

The above parabola corresponds to the function $f(x):=-x^2-4$, but here is my first question: as we need only the "lower half" of the circle for our solution (as can be seen from the picture), is it acceptable to deal only with the function corresponding to that half? In other words, would it be alright just to continue with the function $g(x):=-\sqrt{4-x^2}$?

If the above is OK, then we can continue:
we have:
$f(x)=-x^2-4$,
$g(x)=-\sqrt{4-x^2}$, and we need to find $k, l \in \mathbb{R}$ such that
$t$ ... $y=kx+l$ is a tangent to the graphs $\Gamma_f$ and $\Gamma_g$.

Now, a tangent to $\Gamma_f$ in a point $C(c, f(c))$ will be
$t_f$ ... $y-f(c)=f'(c)(x-c)$.

Likewise, a tangent to $\Gamma_g$ in a point $D(d, g(d))$ will be
$t_g$ ... $y-g(d)=g'(d)(x-d)$.

Or, after some calculation,

$t_f$ ... $y=f'(c)x+f(c)-cf'(c)$

$t_g$ ... $y=g'(d)x+g(d)-dg'(d)$

But we want these two tangents to be one and the same! Therefore, it should be that $t_f = t_g=t$. Since $t$ is expressed as $y=kx+l$, and because equal polynomials have equal coefficients, it follows that

$k= f'(c) =g'(d)$ (1) and
$l=f(c)-cf'(c)=g(d)-dg'(d)$ (2).

When we find $k$ and $l$, the problem will be solved, because the equation of the tangent $y=kx+l$ will be known.

And it is here that I ask for your help, because no matter how I approach this system of two equations in two unknowns,

$-2c=\frac{d}{\sqrt{4-d^2}}$ (which follows from (1))

$-c^2-4-c(-2c)=-\sqrt{4-d^2}-d\frac{d}{\sqrt{4-d^2}}$ (which follows from (2))

it always gets incredibly tangled and when I finally arrive at a polynomial in one variable (say, $c$) it is invariably of the sixth degree!

Or perhaps, the entire procedure described above is flawed? Or is there perhaps a simpler way to find the tangent?

Many thanks!

2. Originally Posted by gusztav
EDIT: After finishing writing this post, it seems to be a bit on the lengthier side, so if you don't have the time to read the whole message, I'm just asking for some help in solving the following system of two equations in two unknowns:

$-2c=\frac{d}{\sqrt{4-d^2}}$

$-c^2-4-c(-2c)=-\sqrt{4-d^2}-d\frac{d}{\sqrt{4-d^2}}$.

But if you're curious how I arrived at this exciting system, read on...

Hi,

I'd be really grateful for some help with this problem:

Find the common tangents to the following curves:
$y+x^2=-4$
$x^2+y^2=4$

When drawn, these curves would look like this:

and our task is to find a line which is tangent to both the circle and the parabola. There will, of course, be two such lines, but that should follow from the calculation.

Let's call the point where the tangent touches the parabola ' $C$', and
the point where the tangent touches the circle ' $D$'.

The above parabola corresponds to the function $f(x):=-x^2-4$, but here is my first question: as we need only the "lower half" of the circle for our solution (as can be seen from the picture), is it acceptable to deal only with the function corresponding to that half? In other words, would it be alright just to continue with the function $g(x):=-\sqrt{4-x^2}$?

If the above is OK, then we can continue:
we have:
$f(x)=-x^2-4$,
$g(x)=-\sqrt{4-x^2}$, and we need to find $k, l \in \mathbb{R}$ such that
$t$ ... $y=kx+l$ is a tangent to the graphs $\Gamma_f$ and $\Gamma_g$.

Now, a tangent to $\Gamma_f$ in a point $C(c, f(c))$ will be
$t_f$ ... $y-f(c)=f'(c)(x-c)$.

Likewise, a tangent to $\Gamma_g$ in a point $D(d, g(d))$ will be
$t_g$ ... $y-g(d)=g'(d)(x-d)$.

Or, after some calculation,

$t_f$ ... $y=f'(c)x+f(c)-cf'(c)$

$t_g$ ... $y=g'(d)x+g(d)-dg'(d)$

But we want these two tangents to be one and the same! Therefore, it should be that $t_f = t_g=t$. Since $t$ is expressed as $y=kx+l$, and because equal polynomials have equal coefficients, it follows that

$k= f'(c) =g'(d)$ (1) and
$l=f(c)-cf'(c)=g(d)-dg'(d)$ (2).

When we find $k$ and $l$, the problem will be solved, because the equation of the tangent $y=kx+l$ will be known.

And it is here that I ask for your help, because no matter how I approach this system of two equations in two unknowns,

$-2c=\frac{d}{\sqrt{4-d^2}}$ (which follows from (1))

$-c^2-4-c(-2c)=-\sqrt{4-d^2}-d\frac{d}{\sqrt{4-d^2}}$ (which follows from (2))

it always gets incredibly tangled and when I finally arrive at a polynomial in one variable (say, $c$) it is invariably of the sixth degree!

Or perhaps, the entire procedure described above is flawed? Or is there perhaps a simpler way to find the tangent?

Many thanks!
Divide equation 1 by -2, so that you have an equation of the form $c = \dots$.

Then substitute this into the second...

3. Originally Posted by Prove It
Divide equation 1 by -2, so that you have an equation of the form $c = \dots$.

Then substitute this into the second...
But that was the first thing I did!

Now, let's see, we get

$c=\frac{-d}{2\sqrt{4-d^2}}$

which should be substituted into

$c^2-4=-\sqrt{4-d^2}-\frac{d^2}{\sqrt{4-d^2}}$

so we get

$\frac{d^2}{4(4-d^2)}-4=-\sqrt{4-d^2}-\frac{d^2}{\sqrt{4-d^2}}$

And what to do next? If I multiplied the whole equation by $\sqrt{4-d^2}$, I'd get

$\frac{d^2}{4 \sqrt{4-d^2}}-4\sqrt{4-d^2}=-4$

... and so on, and eventually I'd arrive at something like $289d^4-1920d^2+3072=0$, and I'm not entirely certain if I'm on the right track here...

This is the reason I have posted the whole problem, maybe the system I got (in $c$, $d$) was wrong in the first place?