# Thread: How do i get the value of (x,y) ?

1. ## How do i get the value of (x,y) ?

Known the following values in the picture below how do i get the value of the x and y. N1N1' and N2N2' are the normals.

I found the first point of contact with the 2nd circle, by solving the line : y=330
and the equation of circle : (x-522)^2 + (y-377)^2 = (130)^2.
(the radii of the 2 circles are 130 each).

Center of Circle c1 (338,377)
Center of Cirlce c2 (522,377)

Even with so much of data i'm unable to find the other point of contact (x,y) with the first circle .......

Wud ne 1 help me out please.............

2. Perhaps if you clarify what the "normals" are?

-Dan

3. Originally Posted by ragsk
Known the following values in the picture below how do i get the value of the x and y. N1N1' and N2N2' are the normals.

I found the first point of contact with the 2nd circle, by solving the line : y=330
and the equation of circle : (x-522)^2 + (y-377)^2 = (130)^2.
(the radii of the 2 circles are 130 each).

Center of Circle c1 (338,377)
Center of Cirlce c2 (522,377)

Even with so much of data i'm unable to find the other point of contact (x,y) with the first circle .......

Wud ne 1 help me out please.............
Zeez, I think your computer's monitor is upside down. In your drawing as posted, the y's are increasing downwards. Umm, cannot be upside down because the x's are increasing rightwards which is normal.
How did you manage to do that? Amazing.

Normal?
Your normals here are lines normal/perpendicular to the circumferences of the two equal circles. So being normal to the circumferences, they pass through the centers of the circles, correspondingly.

So you found the coordinates of the point of normalcy of line N2-N2' to be (400,330).

Actually, it is (400.794,330).
(x-522)^2 +(330-377)^2 = (130)^2
(x-522)^2 = 16,900 -(-47)^2 = 14,691
x-522 = +,-sqrt(14,691) = +,-121.206
x = -121.206 +522 = 400.794 -----------***

Then, the attack could be:
---get the equation of the line passing through the sought-for-(x,y), (400.794,330), and (600,355) points,
---get the intersection of that said line with the circle whose center is at (338,377) to get the (x,y).

----------------------------------
The line passing through (400.794,330), (x,y), (600,355):

slope, m = (355 -330)/(600 -400.794) = 0.1255
Using the point (600,355), the point-slope form of the equation of the said line is
y-355 = (0.1255)(x-600)
y = 0.1255x -75.3 +355
y = 0.1255x +279.7 ----------(1)

-----------------------------------------
The equation of the circle whose center is at (338,377) and whose radius is 130:
(x-338)^2 +(y-377)^2 = (130)^2 -------(2)

The intersection of line (1) and circle (2) is point (x,y).
At that intersection, the coordinates of (1) and (2) are the same, so substitute the y of (1) into (2),
(x-338)^2 +(0.1255x +279.7 -377)^2 = (130)^2
(x-338)^2 +(0.1255x -97.3)^2 = 16,900
Solved the problem. Just simplify, etc.
Here goes,
[x^2 -676x +114,244] +[0.01575x^2 -24.422x +9467.29] -16,900 = 0
1.01575x^2 -700.422x +106,811.29 = 0
Divide both sides by 1.01575,
x^2 -689.561x +105,155.097 = 0
x = {-(-689.561) +,-sqrt[(-689.561)^2 -4(1)(105,155.097)]} /(2*1)
x = {689.561 +,-234.252} /2
x = 461.906 or 227.654
On the drawing, x is to the right of the center of the circle, so,
x = 461.906 -----------***
And, substitute that into (1),
y = 0.1255(461.906) +279.7 = 337.67 ----------***

4. Originally Posted by ticbol
Zeez, I think your computer's monitor is upside down. In your drawing as posted, the y's are increasing downwards. Umm, cannot be upside down because the x's are increasing rightwards which is normal.
How did you manage to do that? Amazing.

Normal?
Your normals here are lines normal/perpendicular to the circumferences of the two equal circles. So being normal to the circumferences, they pass through the centers of the circles, correspondingly.

So you found the coordinates of the point of normalcy of line N2-N2' to be (400,330).

Actually, it is (400.794,330).
(x-522)^2 +(330-377)^2 = (130)^2
(x-522)^2 = 16,900 -(-47)^2 = 14,691
x-522 = +,-sqrt(14,691) = +,-121.206
x = -121.206 +522 = 400.794 -----------***

Then, the attack could be:
---get the equation of the line passing through the sought-for-(x,y), (400.794,330), and (600,355) points,
---get the intersection of that said line with the circle whose center is at (338,377) to get the (x,y).

----------------------------------
The line passing through (400.794,330), (x,y), (600,355):

slope, m = (355 -330)/(600 -400.794) = 0.1255
Using the point (600,355), the point-slope form of the equation of the said line is
y-355 = (0.1255)(x-600)
y = 0.1255x -75.3 +355
y = 0.1255x +279.7 ----------(1)

-----------------------------------------
The equation of the circle whose center is at (338,377) and whose radius is 130:
(x-338)^2 +(y-377)^2 = (130)^2 -------(2)

The intersection of line (1) and circle (2) is point (x,y).
At that intersection, the coordinates of (1) and (2) are the same, so substitute the y of (1) into (2),
(x-338)^2 +(0.1255x +279.7 -377)^2 = (130)^2
(x-338)^2 +(0.1255x -97.3)^2 = 16,900
Solved the problem. Just simplify, etc.
Here goes,
[x^2 -676x +114,244] +[0.01575x^2 -24.422x +9467.29] -16,900 = 0
1.01575x^2 -700.422x +106,811.29 = 0
Divide both sides by 1.01575,
x^2 -689.561x +105,155.097 = 0
x = {-(-689.561) +,-sqrt[(-689.561)^2 -4(1)(105,155.097)]} /(2*1)
x = {689.561 +,-234.252} /2
x = 461.906 or 227.654
On the drawing, x is to the right of the center of the circle, so,
x = 461.906 -----------***
And, substitute that into (1),
y = 0.1255(461.906) +279.7 = 337.67 ----------***

Thanks for the help Ticbol..............
well this works fine only upto a certain value...........
The thing i'm doing here is i'm varying the position of circles keeping the radii and the distance between centres constant...so i'll be getting different centers with different values as the position changes.
Now when the center of the second circle exceeds and i'm getting the point of contact as (0,0)......which is obiviously not on the circle....

Can u please suggest me another method to find the point of contact?

I once again thank u for the pains u've taken to help me in solving...

regards,
ragsk

5. Originally Posted by ragsk
Thanks for the help Ticbol..............
well this works fine only upto a certain value...........
The thing i'm doing here is i'm varying the position of circles keeping the radii and the distance between centres constant...so i'll be getting different centers with different values as the position changes.
Now when the center of the second circle exceeds and i'm getting the point of contact as (0,0)......which is obiviously not on the circle....

Can u please suggest me another method to find the point of contact?

I once again thank u for the pains u've taken to help me in solving...

regards,
ragsk
Keeping the radii (of the two circles?) and the the distance between the two centers constant?
So you just play that same twin cirles entwined that way in different positions (rotations?) with varying coordinates only? (If that is not as confusing as yours, then I don't know how I did it).

The center of the 2nd circle exceeds. Exceeds what? I thought the distance between the two centers, and the radii of both circles, remain constant?

Point of contact of what? The normal line to the first circle? Then how or where or what would be the inclination/position/orientation/... of the normal line? Is the line passing through a known point?

In other words, I cannot fully understand your new/additional question. If you can show a sketch/figure/drawing with given facts/data/constraints/..., then maybe I could help.