Co-ordinate Geometry - Circles & Parabola

Question

A circle with centre at $(15,-3)$ is tangent to $3y=x^2$ at a point in the first quadrant. What is the radius of the circle?

Answer 1

Note that the tangent point is the minimizer of the function $(x,y) \mapsto (x-15)^{2}+(y+3)^{2}$ restricted on the set $\{ (x,y) \mid y = x^{2}/3 \}$. So the problem reduces to minimize the function $x \mapsto (x-15)^{2} + (x^{2}/3 + 3)^{2}$. Could you take it from here?

Answer 2

If the curves are tangent then the slopes are equal at the point of tangency.

Use the derivative to find the slopes of the tangents of both curves.

find $x,y$ such that the tangents are equal.

$(x+3)^2 + (y-15)^2 = r^2\\ 2(x+3) + 2(y-15)\frac {dy}{dx} = 0\\ \frac {dy}{dx} = -\frac {x+3}{y-15}$

$3y =x^2\\ \frac {dy}{dx} = \frac 23 x$

Between:

$-\frac {x+3}{y-15} = \frac 23 x\\ 3y =x^2$

There is enough information to solve your problem.

Answer 3

Let the equation of the circle be $(x-15)^2 + (y+3)^2 = r^2$. The derivative is

$$\begin{align*} 2(x-15) + 2(y+3)\frac{dy}{dx} &= 0\\ \frac{dy}{dx} &= -\frac{x-15}{y+3} \end{align*}$$

The derivative of the parabola is

$$\frac{dy}{dx} = \frac{2x}3$$

Equating the two derivatives give

$$\begin{align*} -\frac{x-15}{y+3} &= \frac{2x}3\\ -3(x-15) &= 2x(y+3)\\ -3x+45 &= 2xy+6x\\ y &= \frac{45-9x}{2x} \end{align*}$$

Substitute $y$ into the parabola $3y= x^2$, (because the tangent point must be on the parabola,)

$$\begin{align*} 3\cdot\frac{45-9x}{2x} &= x^2\\ 135 - 27x &= 2x^3\\ 2x^3+27x - 135 &= 0\\ (x-3)(2x^2+6x + 45) &= 0\\ x &= 3\\ y &= 3\\ r &= \sqrt{(3-15)^2 + (3+3)^2}\\ &= 6\sqrt 5 \end{align*}$$