# Finding a point

• Jun 11th 2013, 04:35 AM
reck1
Finding a point
Greetings. I'm trying to figure out how to find a point on a line, that intersects with a circle. The Radius of my circle is 83.4875 and the 2 points of my line are @ X-81.393 Y2.218 and @ X-84.93 Y.342
The center of the circle and the line points are from X0.0 Y0.0, I've seen some examples online, but not sure how to go about getting my answer. The point I'm looking for is @ X-83.48 Y1.111, can someone show me a step by step on how to get this answer with the information I've given? Many thanks!
• Jun 11th 2013, 04:54 AM
topsquark
Re: Finding a point
Quote:

Originally Posted by reck1
Greetings. I'm trying to figure out how to find a point on a line, that intersects with a circle. The Radius of my circle is 83.4875 and the 2 points of my line are @ X-81.393 Y2.218 and @ X-84.93 Y.342
The center of the circle and the line points are from X0.0 Y0.0, I've seen some examples online, but not sure how to go about getting my answer. The point I'm looking for is @ X-83.48 Y1.111, can someone show me a step by step on how to get this answer with the information I've given? Many thanks!

First you need the equation for your line. You have two points, so find the slope and intercept of your line from this. Next, you know that this line intersects with your circle at some point (x, y). You have two equations: $x^2 + y^2 = r^2 = (83.4875)^2$ and $y = mx + b$. You know m and b so put the y value from your line into the circle formula.

Those are the general steps. See what you can do from here and let us know what, if any, problems you are having.

-Dan
• Jun 11th 2013, 05:41 AM
reck1
Re: Finding a point
So, y=mx+b, "m" is the slope (is that the angle of my line, 27.9412 degrees?), and "b" is the y-intercept (is that my radius, 83.4875?). I'm a little ignorant when it comes to this line of work?!?!
• Jun 11th 2013, 06:03 AM
HallsofIvy
Re: Finding a point
No to both of those. First the "slope", m, of a line is the tangent of the angle the line makes with the x-axis, not the angle itself. That is most easily calculated as $\frac{y_1- y_0}{x_1- x_0}$ which, in this case is $\frac{2.218- 0.342}{81.393- 84.93}$ (you've already done that, just don't take the arctangent!)
To find b, it is better NOT to think of it as the y-intercept. Instead, since you already have m, put that value and either of the two points into y= mx+ b, so that only b is "unknown", and solve for b.
• Jun 11th 2013, 07:09 AM
reck1
Re: Finding a point
I think I'm on the right track, but still coming up empty. I think this is what I need to do: (finding "b") y-mx=b, 2.218-.5304(81.393)=b, so y=.5304(81.393)+40.9528, y=84.1236 (this isn't the answer I was hoping for, any thoughts?
• Jun 11th 2013, 10:23 AM
reck1
Re: Finding a point
Shouldn't the radius of the circle be incorporated into the equation, since that's the intersecting point on the line?
Attachment 28598
• Jun 11th 2013, 11:16 AM
HallsofIvy
Re: Finding a point
"5.305" is NOT the correct slope. I told you before that the slope was given by $\frac{2.218- 0.342}{81.393- 84.93}$. It should be obvious that the numerator is positive while the denominator is negative. The slope here is negative. Do that calculation again.
• Jun 11th 2013, 11:20 AM
HallsofIvy
Re: Finding a point
Quote:

Originally Posted by reck1
Shouldn't the radius of the circle be incorporated into the equation, since that's the intersecting point on the line?
Attachment 28598

Yes, once you have found the equation of the line, in the form y= mx+ b, replace the "y" in the equation of the circle by that so that you have a single quadratic equation in x to solve. A quadratic equation will typically have two solutions giving the x coordinates of the two points of intersection.

(Of course, it is possible that a quadratic equation have no real roots. That is the situation when the line misses the circle completely. Or the equation may have one double root. That is the situation when the line is tangent to the circle.)
• Jun 11th 2013, 11:24 AM
reck1
Re: Finding a point
My bad in the earlier calulations. I finally have a grip on this. Thanks for everyones input!
• Jun 12th 2013, 02:53 AM
reck1
Re: Finding a point
After spending time with the equation, I'm still having issues. I'm getting "b": 2.218- -.5304*-81.393= -40.9528, so y=-.5304*-81.393+ -40.9528=2.218 Going thru that calculation, it gave me the "Y" I started with, I was hoping to get Y1.111, which is the point on my line where the circle intersects. Am I still missing something?
• Jun 12th 2013, 11:30 AM
topsquark
Re: Finding a point
Quote:

Originally Posted by reck1
After spending time with the equation, I'm still having issues. I'm getting "b": 2.218- -.5304*-81.393= -40.9528, so y=-.5304*-81.393+ -40.9528=2.218 Going thru that calculation, it gave me the "Y" I started with, I was hoping to get Y1.111, which is the point on my line where the circle intersects. Am I still missing something?

The equation of your line is y = (-0,5304)x - 40.9528. Now, the point you are looking for is on the line and also the circle. So put this y value into your circle equation:
$x^2 + y^2 = 83.4875^2$

$x^2 + ((-0,5304)x - 40.9528)^2 = 83.4875^2$

and find x.

-Dan
• Jun 12th 2013, 05:22 PM
jpritch422
Re: Finding a point
Actually the slope should be positive:

given the equation of a line is $y = mx + b$ where $m = \frac{y_2 - y_1}{x_2 - x_1}$

$y_2 = 2.218; y_1 = .342; x_2 = -81.393; x_1 = -84.93$

$m = \frac{2.218 - .342}{-81.393 - (-84.93)} = .5304$

the y-intercept is also incorrect. It comes out to be $45.389$

the equation should read $x^2 + (.5304x + 45.389)^2 = (83.4875)^2$