Problem 282. Circle and Quadratic

Created by Matt Fig

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181 Solutions
14 Solvers

Last Solution submitted on Jan 04, 2023

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3 Comments

3 Comments

Gunther S on 14 Feb 2012

the visualisation has an error, it should be

C = @(x) real(-sqrt(R^2-(x-T(1)).^2) + T(2)); % Lower half circle

Matt Fig on 16 Feb 2012

Thanks Gunther. The function does produce the correct graphic, but only when T(1)=0 as in my example.

Rafael S.T. Vieira on 9 Nov 2022

The third test is a pain. Why such a small circle for high precision? It's already a single-point tangent circle. It is no fun to look for the specific method the author had in mind when we already have the answer. IMHO, problems should be open; they accept all kinds of solutions (leaving people to explore clever ideas). Or they should be closed, and a specific method is requested (so people don't wander to dead ends). Precision should be a requirement only for closed ones.

Solution Comments

Solution 9373123

2 Comments

2 Comments

Rafael S.T. Vieira on 9 Nov 2022

r*(x>a) is a slick way to solve the problem for the third test. :)

Christian Schröder on 4 Jan 2023

@Rafael Thanks. :)

Solution 1706613

2 Comments

2 Comments

Christian Schröder on 13 Oct 2022

Actually, the test suite is correct --- try plotting the quadratic curve and the circle from the test suite to see this.

Rafael S.T. Vieira on 9 Nov 2022

I also believe the test suite is incorrect. Using the function roots (after manually calculating the intersection formula), the difference that I got from -4.59997651648516 was 1e-12, which is still greater than the requested 2e-14. William seems to have the right idea (the circle he has found is correct but on the convex side of the parabola).

Solution 34572

3 Comments

3 Comments

Trung Duong on 9 Feb 2012

Is there any special in Test 3?

Austin Senerchia on 5 Apr 2022

x(1) = -b/(2a)
x(2) = (aR^2) - (3/4)*a - (1/4)(b^2/a) + c

why is this not right? it's close

Rafael S.T. Vieira on 9 Nov 2022

I had to make a special test only for the 3rd test because it's a single-point tangent circle. A pebble so small it seems to be resting on a nearly flat surface.

Solution 34498

2 Comments

2 Comments

Trung Duong on 9 Feb 2012

don't know why the assertions fail.
The solution is correct.

Matt Fig on 9 Feb 2012

If you look at the example posted in the problem description, you will see that the return argument should be a two element row vector.

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Problem 282. Circle and Quadratic

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Problem 282. Circle and Quadratic

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