Finding all tangent lines to circle $x^2+y^2=9$ having exactly one common point with parabola $y=x^2+3$

Question

I have to find all tangent lines to the circle $x^2+y^2=9$, which have exactly one common point with the parabola $y=x^2+3$.

I drew it and I saw that $(0,3)$ is a common point circle and parabola. I found that three tangent lines are: $y=3$, $x=-3$, $x=3$. And now the question is whether these are all? I don't know how to explain it or find others lines.

Thanks for your help!

Answer 1

At any given point on the circle, there is a unique tangent line. At the point $(0,3)$ you have identified 3. Only one is tangent.

The other lines:

Since this is tagged pre-calculus, I will not use calculus in this answer.

Suppose $x_1,y_1$ lies on the circle.

Slope of the tangent line though that point is negative reciprocal of the slope of the radius.

$m = -\frac {x_1}{y_1}$

Equations of the tangent line in point slope and y-intercept form:

$y-y_1 = -\frac {x_1}{y_1} (x-x_1)\\ y = -\frac {x_1}{y_1} x + \frac {x_1^2 +y_1^2}{y_1}\\ y = -\frac {x_1}{y_1} x + \frac {9}{y_1}\\ $

Remember $x_1,y_1$ lie on the circle so $x_1^2 + y_1^2 = 9$ we will be using this fact a couple more times.

The line is also tangent to $y = x^2 + 3$

Substituting:

$-\frac {x_1}{y_1} x + \frac {9}{y_1} = x^2 + 3\\ y_1x^2 + x_1x + 3y_1 - 9 = 0$

Since our line is tangent, when we plug this into the quadratic formula, we must get exactly one root. i.e. $b^2 - 4ac = 0$.

$x_1^2 - 4y_1(3y_1-9) = 0\\ x_1^2 + 36y_1 - 12y_1^2 = 0$

We can use $x_1^2+ y_1^2 = 9$ again. Substitute $x_1^2 = 9 - y_1^2$

$(9-y_1^2) + 36y_1 - 12y_1^2 = 0\\ 13y_1^2 - 36y_1 - 9 = 0\\ (y_1 - 3)(13y_1 + 3) = 0$

We have two values for $y_1.$ Solve for $x_1.$

$x_1^2 + 9 = 9\\ x_1^2 + \frac {9}{169} = 9$

$(0,3),(\frac {3\sqrt {168}}{13},-\frac {3}{13}), (-\frac {3\sqrt {168}}{13},-\frac {3}{13})$ are our 3 points of tangency.

And, plug into our earlier equations of the tangent lines.

$y = 3\\ y = \sqrt {168} x - 39\\ y = -\sqrt {168} x - 39$

Answer 2

Any point on the parabola : $P(a,a^2+3)$

The equation of the tangent at $P$ will be $$\dfrac{y-(a^2+3)}{x-a}=2a$$

$$\iff2ax-y+3-a^2=0$$

To be a tangent of the circle, the distance from the centre needs to be equal to the radius

$$3=\dfrac{|3-a^2|}{\sqrt{(2a)^2+(-1)^2}}$$

Take square in both sides and solve for $a$

Answer 3

Let $y = ax+b$ be such a line. A pair $(a, b)$ forms a solution if the corresponding line intersects both the parabola and the circle in exactly one point each. So, substituting $y$ in each curve equation:

$(ax+b)^2 + x^2 = 9$ must have only one solution: discriminant, in terms of a and b, is zero.

$(ax+b)= x^2 +3$ must have only one solution: discriminant, in terms of a and b, is zero.

You then get a system of equations, which is very straightforward:

$9a^2=b^2-9$

$a^2=4(3-b)$

Observe that, if $(a, b)$ is a solution, so is $(-a, b)$, as the curves are geometrically simmetric towards y axis.

EDIT

Vertical lines are not checked for solution in the former assumption that $y = ax+b$. Setting $x = k$, constant, should find the other lines you got from the beginning.