Find the radius of circle which touches parabola and $x$-axis

Question

I am working on my scholarship exam practice (high school/pre-university maths) and attempt to do this question but stuck.

Denote by $D$ the Domain

$\{(x,y)|x\geq0, y\geq0\}$.

Assume that a circle $C$ contained in $D$ touches the parabola $y=\frac{1}{2}x^2$ at the point $(2,2)$ and also touches the $x$-axis. Find the radius of $C$.

I am not sure how I should interpret this problem. What I know so far is below:

I have wrote a rough graph below, but I think I should consider only quadrant 1 because of the domain.

For the circle if we assume that circle may or may not be centered at origin, then

$(x-a)^2+(y-b)^2=r^2$

$(2-a)^2+(2-b)^2=r^2$ at $(2,2)$

It also touches $x$-axis, meaning that

$(r-a)^2+(b^2)=r^2$

$(2-a)^2+(2-b)^2=(r-a)^2+(b^2)$

After factorized and rearranged, we got

$a=\frac{r^2-8}{2(r-4)}$

And I am not sure what to do from here, I am not even sure that I am on the right track. The answer key to this question is $\frac{5-\sqrt{5}}{2}$. Please help.

Answer 1

As stated in the comment, "touching" and "intersecting" has different meanings. "Touching" at point $P$ implies for both figure should pass through $P$ and their tangents at $P$ should be parallel or coincident.

Here, given equation of Parabola $$P: y = \cfrac{x^2}{2}$$ $$ \Rightarrow y_1'(x,y) = x$$ General equation of circle $C$ which is touching $x$ - axis is $$C: (x-a)^2 + (y-b)^2 = b^2 $$ $$ \Rightarrow y'_2(x,y) = - \cfrac{x-a}{y-b}$$

For touching at $(2,2)$, $y_1'(2,2) = y_2'(2,2)$ $ \Rightarrow a + 2b = 6$

For intersection, $C$ should pass through $(2,2)$, put $x = 2, y = 2$ in the equation of same, you will get the equation in terms of $a$ and $b$, now you have two equation and two unknown. I guess, you can do it from here.

After solving $a = 1 \pm \sqrt{5}, b = \cfrac{5 \mp \sqrt{5}}{2}$ As you can see that only the red circle satisfy the domain criterion, hence $ b = \cfrac{5 - \sqrt{5}}{2} $ , which is the radius of the circle.

Answer 2

Without calculus:

If the circle touches the $x$-axis, then its radius is equal to the absolute value of the $y$-coordinate of its center. So, if $(a,b)$ are the coordinates of its center, an equation of such a circle is $$(x-a)^2+(y-b)^2=b^2.\tag1$$

“Touching” here means that the circle and parabola have a common tangent at their intersection point. If you have a conic with general equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, you can obtain the equation of its tangent at the point $(x_0,y_0)$ via a set of substitutions: $$x^2\to xx_0 \\ xy \to \frac12(xx_0+yy_0) \\ y^2\to yy_0 \\ x\to\frac12(x+x_0) \\ y\to\frac12(y+y_0).$$ So, an equation of the tangent to the parabola at $(2,2)$ is $$x-\frac12(y+2)=0\tag2$$ and an equation of the tangent to the circle is $$2x-a(x+2)+2y-b(y+2)+a^2 = 0.\tag3$$ For these equations to represent the same line, the coefficients of one must be multiples of the corresponding coefficients of the other. Rearranging equations (2) and (3) and comparing coefficients generates the system $$2-a=k \\ 2-b=-\frac12k \\ a^2-2a-2b=-k,$$ which you should be able to solve for $a$ and $b$.

Alternatively, the center of the circle must lie on the perpendicular to the tangent at $(2,2)$. From equation (2), the slope of the tangent at this point is $2$, so an equation of the perpendicular to this tangent is $${y-2\over x-2}=-\frac12.\tag4$$ Additionally, given two tangents to a circle, the circle’s center lies on one of the angle bisectors of those tangents. An equation of the $x$-axis is $y=0$, so the two bisectors of your circle have equations $$\frac1{\sqrt5}(2x-y-2)=\pm y.\tag5$$ Compute the intersections of line (4) with each of the lines in (5).

Another way is similar to how you started to solve the problem. Since $(2,2)$ satisfies $(1)$, we must have $(2-a)^2+(2-b)^2=b^2$. The coordinates of the center must also satisfy equation (4). Solve the resulting system of two equations in $a$ and $b$.

With any of these methods, you will end up with two potential solutions. Reject the solution that doesn’t lie in $D$.

Answer 3

The circle with center $a,b$ tangent to $x$-axis means that the radius will be the same as $b$ because tangent in circle will always perpendicular. A line which goes from center to $(2,2)$ will have gradient $-1/2$, while distance from center to $(2,2)$ is radius.