# iOSfind my point if inside a circle.

#### chhoda

##### macrumors 6502
Original poster
I have two cgpoints say pt1 pt2

pt2 is center of a circle.

how can i find if pt1 is within the perifery of a 50px radius circle centered at pt2 ?

ch

#### jonnymo5

##### macrumors 6502
Sounds like you need the distance formula.

Go to www.google.com and type this into the search bar. "what is the distance between two points"

#### mobilehaathi

##### macrumors G3
Just check if they are within 50px of each other.

#### ArtOfWarfare

##### macrumors G3
Are you familiar with

A^2 = B^2 + C^2?

A is 50
B is the difference in X's between your two points.
C is the difference in Y's between your two points.

The formula defines the boundary of your point. So you'd actually want to use
A^2 > B^2 + C^2
to find out if your two points were within 50 of each other.

#### dejo

##### Moderator
Staff member
Are you familiar with

A^2 = B^2 + C^2?
No. But I am familiar with A^2 + B^2 = C^2. So you'd actually want to use
A^2 > B^2 + C^2
to find out if your two points were within 50 of each other.
I'd think something like:
Code:
``if (sqrt(b*b + c*c) <= 50.0)``
would be more clear.

#### ArtOfWarfare

##### macrumors G3
I may be running on a shortage of sleep today...

xD

#### maril1111

##### macrumors 68000
A^2 + B^2 = C^2[/URL].
This is probably one of the first times i see this formula being applied and used in real life, i always thought its just another pointless topic that the school forces you to learn.

#### mobilehaathi

##### macrumors G3
I'd think something like:
Code:
``if (sqrt(b*b + c*c) <= 50.0)``
would be more clear.
More clear, yes, and hopefully the OP understands now. However, wouldn't a sqrt function be rather expensive compared to an extra multiplication? Of course that only matters if the check is happening frequently.

#### ArtOfWarfare

##### macrumors G3
This is probably one of the first times i see this formula being applied and used in real life, i always thought its just another pointless topic that the school forces you to learn.
Math is meaningless without programming.

(Ok, fine, the other sciences give meaning to math as well...)

Edit: And math team. Somehow math team gives math purpose.

#### dejo

##### Moderator
Staff member
However, wouldn't a sqrt function be rather expensive compared to an extra multiplication?
I don't know. Do you?

#### ArtOfWarfare

##### macrumors G3
I made a quick app that tests which is faster, this is the output:
Code:
``````2012-01-30 13:13:53.915 SquareRootCost3[22875:f803] Beginning Test with squares.
2012-01-30 13:13:53.941 SquareRootCost3[22875:f803] Finished test with squares - found 369212
2012-01-30 13:13:53.941 SquareRootCost3[22875:f803] Beggining test with roots.
2012-01-30 13:13:53.954 SquareRootCost3[22875:f803] Finished test with roots - found 369212``````
So the squares took 0.026 seconds while the roots took 0.013 seconds... it appears to take half as long to use squares. Odd.

My code, for anyone interested, is this:
Code:
``````    srand(time(NULL));

int i = 0;
float a, b, c;
for (i = 0; i < 500000; i++)
{
a[i] = rand();
b[i] = rand();
c[i] = rand();
}

long int squareans = 0;
long int rootans = 0;

NSLog(@"Beginning Test with squares.");
for (int i = 0; i < 500000; i++)
{
if (a[i]*a[i] < (b[i]*b[i] + c[i]*c[i]))
{
squareans++;
}
}

NSLog(@"Finished test with squares - found %ld", squareans);
NSLog(@"Beggining test with roots.");
for (int i = 0; i < 500000; i++)
{
if (a[i] < sqrtf(b[i]*b[i] + c[i]*c[i]))
{
rootans++;
}
}

NSLog(@"Finished test with roots - found %ld", rootans);``````

#### dejo

##### Moderator
Staff member
sqrt() can probably take advantage of bitwise operators.

#### mobilehaathi

##### macrumors G3
I made a quick app that tests which is faster, this is the output:
Code:
``````2012-01-30 13:13:53.915 SquareRootCost3[22875:f803] Beginning Test with squares.
2012-01-30 13:13:53.941 SquareRootCost3[22875:f803] Finished test with squares - found 369212
2012-01-30 13:13:53.941 SquareRootCost3[22875:f803] Beggining test with roots.
2012-01-30 13:13:53.954 SquareRootCost3[22875:f803] Finished test with roots - found 369212``````
So the squares took 0.026 seconds while the roots took 0.013 seconds... it appears to take half as long to use squares. Odd.
Cool, that was my intuition from taking a numerical analysis course long ago, but thanks for demonstrating it. ----------

sqrt() can probably take advantage of bitwise operators.
Hmm, it will probably change based on implementation. What is the difference between sqrt() and sqrtf()?

Edit: Oh it is for floats. It'll probably be slower than one optimized for ints.

Edit2: Ok, not quite. sqrtf() explicitly takes floats, where as sqrt() is overloaded and can take floats or doubles (or other stuff in C++).

Last edited: