# Tricky Question - challenge

• Dec 5th 2006, 07:38 PM
thejoester
Tricky Question - challenge
my teacher said anyone in my class who gets this question would get a prize in front of the whole school.... okay.... i exaggerate.... but still, i want to understand this.

the question is:

consider the circle x^2 + y^2 = 25 and the line y = x +k, where k is any real number. Determine the values of k for which the line will intersect the circle in one, two, or no points. Repeat for the circle x^2 + y^2 = 49 and the line y = x+k.

generalize your results for a circle of radius r and the line y = x+k

Note: the question asks for you to do two. Dont worry about it, just explain the steps and i will understand it.

• Dec 5th 2006, 07:57 PM
ThePerfectHacker
Quote:

Originally Posted by thejoester
consider the circle x^2 + y^2 = 25 and the line y = x +k, where k is any real number. Determine the values of k for which the line will intersect the circle in one, two, or no points. Repeat for the circle x^2 + y^2 = 49 and the line y = x+k.

What class is this?

Okay,
If they intersect we can substitute $\displaystyle y=x+k$ into the equation,
$\displaystyle x^2+(x+k)^2=25$
$\displaystyle x^2+x^2+2xk+k^2=25$
$\displaystyle 2x^2+2xk+k^2=25$
$\displaystyle 2x^2+2xk+(k^2-25)=0$
To have 2 distinct points we need the discrimant to be positive.
$\displaystyle (2k)^2-4(2)(k^2-25)>0$
Thus,
$\displaystyle 4k^2-8k^2+200>0$
$\displaystyle 200-4k^2>0$
$\displaystyle 50-k^2>0$
$\displaystyle 50>k^2$
$\displaystyle k^2<50$
$\displaystyle |k|<\sqrt{50}$
• Dec 6th 2006, 08:37 AM
Soroban
Hello, thejoester!

Quote:

Consider the circle $\displaystyle x^2 + y^2 \,= \,25$ and the line $\displaystyle y \,= \,x +k$
Determine the values of $\displaystyle k$ for which the line will intersect the circle in 1, 2, or 0 points.

To complete what ThePerfectHacker started . . .

The discriminant is: .$\displaystyle D \:=\:4(50 - k^2)$

To have one point of intersection, $\displaystyle D = 0$
. . $\displaystyle 4(50 - k^2)\:=\:0\quad\Rightarrow\quad k^2 \,=\,50\quad\Rightarrow\quad k \,=\,\pm5\sqrt{2}$

To have two points of intersection: $\displaystyle D > 0$
. . $\displaystyle 4(50-k^2)\:>\:0\quad\Rightarrow\quad k^2 \:<\:50\quad\Rightarrow\quad |k| \:< \:5\sqrt{2}$

To have no points of intersection: $\displaystyle D < 0$
. . $\displaystyle 4(50-k^2) \:<\:0\quad\Rightarrow\quad k^2 \:>\:50\quad\Rightarrow\quad|k|\:>\:5\sqrt{2}$

The "graph" looks like this:

. . $\displaystyle \overbrace{- - - }^{\text{0 points}}+\overbrace{ - - - + - - - }^{\text{2 points}}+\overbrace{ - - -}^{\text{0 points}}$
. . . . . $\displaystyle \text{-}5\sqrt{2}$ . . . . . . . . .$\displaystyle 5\sqrt{2}$
. . . . . $\displaystyle ^{\text{1 point}}$ . . . . . . . . .$\displaystyle ^{\text{1 point}}$

• Dec 6th 2006, 02:28 PM
thejoester
Thanks!
thanks a lot guys, i really appreciate it.
i have one more question though, would either of you know how to graph this on excel?