Find center of circle inscribed by 2 curves

A

AvgStudent

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Another challenge from my professor.
Find the centre of the largest circle bounded by the curves: [imath]y=\sqrt{x}[/imath] and [imath]y=x^2[/imath]
Screen Shot 2022-02-13 at 2.45.24 AM.png
(Graph by Desmos)
I think the centre will lie on the line [imath]y=x[/imath] (in purple). Brainstorming...?
 
AvgStudent said:
Another challenge from my professor.
Find the centre of the largest circle bounded by the curves: [imath]y=\sqrt{x}[/imath] and [imath]y=x^2[/imath]
View attachment 31089
(Graph by Desmos)
I think the centre will lie on the line [imath]y=x[/imath] (in purple). Brainstorming...?
Click to expand...
I agree with your guess. In fact, I believe you can actually prove this conjecture.
 
Think about a circle that touches both curves. Sketch a tangent to each curve at the 2 touching points. Sketch a tangent to the circle the points. Any observations? What will the gradient of the tangents be when the circle is largest?
 
My initial thought is that the largest circle will have the largest diameter, i.e. the max distance between the two curves.
Let [imath](x_1,\sqrt{x_1})[/imath] and [imath](x_2,x_2^2)[/imath] be points on the curves. Therefore, I need to maximize the squared distance (easier to take the derivative than the distance):
[math]D^2=(x_1-x_2)^2+(\sqrt{x_1}-x_2^2)[/math]But I'm not sure if I can take the derivative respect to [imath]x[/imath] since I have [imath]x_1 \, \&\,x_2[/imath]. Even then, I have one equation with two unknowns.
Cubist said:
Think about a circle that touches both curves. Sketch a tangent to each curve at the 2 touching points.
Click to expand...
The two tangent curves will be parallel, and the circle's centre will be perpendicular to the tangents.
I'm not sure I understand this sentence: "Sketch a tangent to the circle the points."?
 
AvgStudent said:
I'm not sure I understand this sentence: "Sketch a tangent to the circle the points."?
Click to expand...
Sorry, I meant to write, "sketch a tangent to the circle at the intersection points."

Consider the two red curves sketched below. A touching circle is sketched in green. The two tangents are drawn in black.

p1.png
At one intersection point, say the top one, the tangent to the circle and the tangent to the curve are exactly the same therefore only one tangent line is drawn.

Consider the two tangents for the two separate points. If they are different gradients, like they are in the sketch, they will form a wedge and the circle could increase in size if its centre is moved (in the sketch it could move to the right and the circle could increase in size). If you can prove that the circle's centre lies on y=x, then what would the gradient of the tangents be when the circle is biggest?
 
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Cubist said:
Sorry, I meant to write, "sketch a tangent to the circle at the intersection points."

Consider the two red curves sketched below. A touching circle is sketched in green. The two tangents are drawn in black.

View attachment 31104
At one intersection point, say the top one, the tangent to the circle and the tangent to the curve are exactly the same therefore only one tangent line is drawn.

Consider the two tangents for the two separate points. If they are different gradients, like they are in the sketch, they will form a wedge and the circle could increase in size if its centre is moved (in the sketch it could move to the right and the circle could increase in size). If you can prove that the circle's centre lies on y=x, then what would the gradient of the tangents be when the circle is biggest?
Click to expand...
Could not resist some nitpicking: while not true for this problem, in a general case the largest tangential circle might not be bounded by the curves.
 
Thank you @Cubist, for the beautiful picture. I forgot to state the domain [imath]x \in [0,1][/imath]
Can I argue that due to symmetry, and in addition to the fact that the tangent lines of the intersections of the circle and both curves must have the same gradient as y=x?
 
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blamocur said:
Could not resist some nitpicking: while not true for this problem, in a general case the largest tangential circle might not be bounded by the curves.
Click to expand...
Indeed. I was hoping that AvgStudent might spot this too.

AvgStudent said:
Thank you @Cubist, for the beautiful picture. I forgot to state the domain [imath]x \in [0,1][/imath]
Can I argue that due to symmetry, and in addition to the fact that the tangent lines of the intersections of the circle and both curves must have the same gradient as y=x?
Click to expand...
Yes.

You might also want to check that each curve only touches the circle at one point only to address the point that @blamocur points out. For example, see the pic below which shows a situation that would require a different method...
p2.png
 
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Sorry for the late response. As this is a challenge problem, I only spend time on it when I get some free time.
Since the gradient of [imath]y=x[/imath] is 1, then we can find the intersecting point [imath]x[/imath] by setting the gradient of [imath]x^2[/imath] i.e [imath]2x=1\implies x=\frac{1}{2}[/imath]
Similarly, setting the gradient of [imath]\sqrt{x}[/imath] i.e. [imath]\frac{1}{2\sqrt{x}}=1 \implies x=\frac{1}{4}[/imath].
The midpoint between the two x-values is 3/8. Is this the answer?
 
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Consider all lines segments perpendicular to the line y=x that go from y= sqrt(x) to y=x^2.
Of all these line segments, which one is longest?
I agree with your claim that this longest segment will be the diameter of the circle. But remember that this diameter must be perpendicular to the line y=x
Now go and find it.
 
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