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There is a circle with a radius of $25$ ft and origin at $(0, 0)$ and a line segment from (0, -31) to (-37, 8). Find the intersections of the line and circle.

I am asking for somebody to analyze what I am doing wrong in calculating the answer, given the question above in its exact format. I almost never get an answer correct, and I would like an explanation as to why this appears to be the case.

I begin with finding the formula of the line

I used the slope formula, $\Delta y \over{\Delta x}$ to get ${-31 - 8 \over{0 - (-37)}} = {-39 \over{37}}$, which becomes $1.054$, so that $y = 1.054x + b$
$(0, -31)$ is the y-intercept when $x = 0$, so $y = 1.054x - 31$.

Formula of circle

The format I learned is $(x - h)^2 + (y - k)^2 = r^2$.
Substitute known info: $(x - 0)^2 + (y - 0)^2 = 25$
Simplify: $x^2 + y^2 = 25$

Substitution

I took the circle's equation $x^2 + y^2 = 25$ and plugged in the value of y: $x^2 + (0.621x - 31)^2 = 25$.
I expanded and rounded it into $x^2 + 1.111x^2 + -65.348x + 961 = 25$.
It simplifies to $2.111x^2 - 65.348x + 936$.

Quadratic formula

The template: $x = {-b \pm \sqrt{b^2 - 4ac }\over{2a}}$ is used by plugging in values:
${65.348 \pm \sqrt{4270.361 - 4*2.111*936 }\over{2*2.111}} = {65.348 \pm \sqrt{4270.361 - 7903.584}\over{4.222}} = {65.348 \pm \sqrt{-3633.223}\over{4.222}}$.

Either I do not know what to do next or (as I suspect) this is the unfortunate answer to the problem. The problem continues with questions about how long somebody is in the circle if they follow the line at a certain rate, which leads me to believe that there are real roots, but I certainly cannot find them. Edit: slope and template errors corrected, answer is still incorrect.

Caleb Stanford
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kettlecrab
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    When you got the negative slope of $-23/37$ you could have kept it as a fraction. But when you converted it to a decimal it became positive, which it should not be. – coffeemath Oct 27 '13 at 21:54
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    Actually your initial calculation of the slope is wrong even before you drop the sign. It should be $[(-31)-(8)]/[0-(-37)]=-39/37$ which is very close to 1. If the slope calculation is wrong you get a line which does not intersect the circle. – coffeemath Oct 27 '13 at 22:07
  • Both issues now fixed; thank you. – kettlecrab Oct 27 '13 at 22:26
  • You fixed the slope in the formula for the line, but kept the old formula when plugging into the circle equation. Actually you still have $+1.054x-31$ It should be $-1.054x-31$ in all places, if you're going to approximate things. – coffeemath Oct 27 '13 at 22:40

3 Answers3

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You made a mistake in citing the Quadratic Formula. The 'template' you need is:

$$x= \frac{-b\pm \sqrt{b^2 -4ac}}{2a}$$

Now you should be able to solve it.

Newb
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Check your slope again: It is $-\frac{39}{37}\approx-1.05$. Check your circle, it is $$x^2+y^2=625,$$ that is, $r^2=25^2=625$. I think you can get it from there, although I recommend you use fractions throughout until the end. For my final gift, here is a plot. If you hover over the intersection points, it will tell you what they are so you can verify your work.

J. W. Perry
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  • There are questions that follow this on my homework, such as (and this is assuming that there is a person walking along the line at a rate of 5 ft/s), "Where did Jerry enter and leave the circle?" as well as "On Jerry's walk, how much time was he inside the circle?" There are four of these questions, and it is pretty obvious that the line is supposed to intersect (at least, I would think so). I fixed the slope. – kettlecrab Oct 27 '13 at 22:30
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    Aha I see another error, and yes they probably do intersect. Your circle equation, not 25 but $25^2$ – J. W. Perry Oct 27 '13 at 22:43
  • And I deliberately left it as 25 because I thought that it was meaning that it was the answer (now that I read it again, it was the radius and must be squared). – kettlecrab Oct 27 '13 at 23:25
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The line equation is $y=(-39/37)x-31.$ To get the closest point on this line to the origin, intersect it with the perpendicular to it from $(0,0),$ which has equation $y=(37/39)x$. The nearest point on the line to $(0,0)$ is then seen to be $(a,b)$ where $a=-44733/2890,\ b=-42439/2890.$ The distance to the origin is then $\sqrt{a^2+b^2} \approx 21.336.$ Since this is less than the required radius of $25$ there should be two points for the solution. [I could have sworn that initially the given radius was 5...]

coffeemath
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    The given radius has always been 25 (and is supposed to be $x^2 + y^2 = 25^2$), as stated in the original problem. I made the mistake of thinking it was $5^2$ as well. – kettlecrab Oct 27 '13 at 23:34