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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.

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    $\begingroup$ To "touch" and "intersect" has different meanings,In your graph circle and parabola are intersecting $\endgroup$ – Ajay Mishra Jun 18 at 11:51
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    $\begingroup$ In this context, touching means it is tangent. So the circle is tangent to the $x$ axis$ and to the parabola. $\endgroup$ – Andrei Jun 18 at 11:51
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    $\begingroup$ Your picture does not describe your text. Isn't the circle supposed to be inside the domain $D$? $\endgroup$ – uniquesolution Jun 18 at 11:51
  • $\begingroup$ Thank you all for the heads up, let me take some time reattempting this question. :) $\endgroup$ – Trey Anupong Jun 18 at 12:10
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enter image description hereAs 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.

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  • $\begingroup$ Hi, I am trying to find $y'_2(x,y)$, I got $y'_2(x,y)=\frac{1-2(x-a)}{2(y-b)}$ by differentiating general equation of circle $C$ to $2(x-a)+2(y-b)\frac{dy}{dx}=1$, how did you come up with $\frac{x-a}{x-b}$ above? $\endgroup$ – Trey Anupong Jun 18 at 12:27
  • $\begingroup$ That is $- \cfrac{x-a}{y-b}$ , I forgot the minus sign in the answer(Thanks!, I edited that, but the answer won't change, as I had incorporated that in my solution) $\endgroup$ – Ajay Mishra Jun 18 at 12:41
  • $\begingroup$ Equation: $x^2 + a^2 - 2xa + y^2 - 2by = 0$ on differentiating w.r.t to $x$, $ 2 x + 0 - 2a + 2y y' - b y' = 0$ on dividing both sides by 2, $ (x-a) + y' ( y - b) = 0 \Rightarrow y' = - \cfrac{x-a}{y-b}$ $\endgroup$ – Ajay Mishra Jun 18 at 12:44
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    $\begingroup$ If you have problem with this form, then proceeding via your approach $ 2(x-a) + 2(y-b) y' = 0 \Rightarrow (x-a) + (y-b) y' = 0 \Rightarrow y' = -\frac{x-a}{y-b} $ $\endgroup$ – Ajay Mishra Jun 18 at 12:45
  • $\begingroup$ You got that now? $\endgroup$ – Ajay Mishra Jun 18 at 12:46
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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 $D$.

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