# Finding the points of intersection of a circle and a line

In a test (of math in arabic language) we we're asked to find the points of intersection of a circle and a line. Their equation is given.

In the test I solved system of equations made of their equation and in the process I explain my line of thought using the words therefore, then, so, ... etc (in arabic) but in describing that process I acknowledge that the structure of my proof is done by equivalence, that can be easily seen from the context, we're solving an equation so we proceed by equivalence.

But my professor said that it was all wrong and that I should have used the statement is equivalent (in arabic) in each part of my proof and I got all of my exercises wrong, and there were some exercises in the test where in the process we had to solve an equation so he also said that I was wrong because of that reason so I only got 4 out of 20.

Is the professor wrong or I am?

My computer crashed so i couldn't edit so i made a new question. I've added the equations and system of equations with my solutions

Exercise 1

Circle has equation $x^2+y^2+(m+2)x-2my+m^2-36=0$, find center, radius

I found it to be center $(-\frac{m}{2}-1,m)$ radius r= $\sqrt{\frac{m^2}{4}+m+37}$

Exercise 2

Circle has equation $x^2+y^2+2 x - 2y - 4 = 0$ line has equation $x+y-1=0$ find points of intersection

i found $\left(\frac{-1+\sqrt{11}}{2},\frac{3-\sqrt{11}}{2}\right)$ and $\left(\frac{-1-\sqrt{11}}{2},\frac{3+\sqrt{11}}{2}\right)$

for exercise 3 and 4 it is the same and i checked my calculations again and again with the calculator there's no error but yet all of that was marked with 0 for not saying "is equivalent", i didn't use the symbol $\Rightarrow$ All i got was 4 from exercise 5 which had calculation of dot product and of area of triangle where we are given coordinates of two vectors in the plane

• In Exercise 2, your two points do not satisfy $x + y - 1 = 0$. – dfan Dec 19 '14 at 20:28
• You should have edited: math.stackexchange.com/questions/1074894/… instead of making a duplicate post. Also, why did you start a new user account since you must be the same user as the other post? – dustin Dec 19 '14 at 20:36
• this is comment: i made a typo it should be $$(\frac{-1-\sqrt{11}}{2},\frac{3+\sqrt{11}}{2})$$ and $$(\frac{-1+\sqrt{11}}{2},\frac{3-\sqrt{11}}{2})$$ sorry again my computer crashed i made an edit but it should be approved – ecaps ro Dec 19 '14 at 20:42
• This was cross posted at matheducators.stackexchange.com/questions/6056/… – Chris C Dec 19 '14 at 22:03

The posted solutions are correct, below is a possible derivation.

Exercise 1

In this exercise you need to find the center and radius of this circle which depends on $m$.

$$x^2+y^2+(m+2)x-2my+m^2-36=0$$

Notice that $(y-m)^2=y^2+m^2-2my$ and rewrite as

$$x^2+(y-m)^2+(m+2)x-36=0$$

Add $0=1-1+m+\frac{m^2}4-m-\frac{m^2}4$

$$x^2+(y-m)^2+(m+2)x-37+1+m+\frac{m^2}4-m-\frac{m^2}4=0$$

Notice that $\left(x+\frac{m-2}2\right)^2=x^2+(m+2)x+\frac{m^2}4+m+1$ and rewrite as

$$\left(x+\frac{m+2}2\right)^2+(y-m)^2=37+m+\frac{m^2}4=0$$

We can now easily see that the center is

$$\left(-\frac{m+2}2,m\right)$$

$$r=\sqrt{37+m+\frac{m^2}4}=\frac12\sqrt{m^2+4m+148}$$

Which is equivalent to your solution.

Exercise 2

Circle has equation line has equation find points of intersection

You need to find the intersections of the following:

$$\begin{array}{l} 0=x^2+y^2+2 x - 2y - 4\\ 0=x+y-1 \end{array}$$

This simply means solving a system of equations:

From the second equation we get the solution for $x$ as

$$x=1-y$$

Now plug this into the first equation:

$$0=(1-y)^2+y^2+2 (1-y) - 2y - 4$$

Remove the paranthesises

$$0=2y^2-6y - 1$$

Solve for $y$ using the quadratic formula or complete the square

$$y_1=\frac12 (3+ \sqrt{11})$$ $$y_2=\frac12 (3- \sqrt{11})$$

Plug this into the formula for $x$ from earlier: $x=1-y$

$$x_1=1-\frac12 (3+ \sqrt{11})$$ $$x_2=1-\frac12 (3- \sqrt{11})$$

Simplify $$\begin{array}l x_1=-\frac12(1+\sqrt{11})\\ x_2=\frac12(\sqrt{11}-1) \end{array}$$

This gives the two solutions:

$$\left(-\frac12(1+\sqrt{11}),\frac12 (3+ \sqrt{11})\right)$$ and $$\left(\frac12(\sqrt{11}-1),\frac12 (3- \sqrt{11})\right)$$

Which is equivalent to the posted solutions.