Challenging Question - help a brother

**thejoester** · December 9th 2006, 06:39 AM

Hello,
thanks in advance for anyone who can help me.

this question relates to linear quadratic systems

the question is:

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

I thought maybe to help, I would put the question before this one, which relates to this question.
It goes like this

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.

I figured out how to do this by subbing, x+k into the other equation, and then using discriminants to find out the values of k that would work.

im confused on the generalization question though.

Again, thanks to anyone who can help me

**Plato** · December 9th 2006, 08:07 AM

You really should graph this problem.
The line $y = x + k$ is tangent to $x^2 + y^2 = {25}$
at $\left( { - \sqrt {12.5} ,\sqrt {12.5} } \right)\quad \& \quad \left( {\sqrt {12.5} , - \sqrt {12.5} } \right).$

If you use those points to find values of k and if you graph this correctly, you will see the answer.

**earboth** · December 9th 2006, 08:09 AM

Originally Posted by thejoester

...

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

...

Hello,

the circle has the equation:
$x^2+y^2=r^2$ and the line has the equation: $y=x+k$ .

To calculate the intercepts you plug in the term of the line into the equation of the xircle:
$x^2+(x+k)^2=r^2$ . After a few steps you'll get:

$2x^2+2kx+k^2-r^2=0$ . Solve for x. Use the formula to solve quadratic equations:

$x=\frac{-2k \pm \sqrt{4k^2-4 \cdot 2 \cdot (k^2-r^2)}}{2 \cdot 2}=$ $\frac{-2k \pm \sqrt{8r^2-4k^2}}{4}$ = $\frac{-k \pm \sqrt{2r^2-k^2}}{2}$

You get 2 intercepts if the discriminant is greater than zero: ${2r^2-k^2}>0$
You get 1 intercept if the discriminant equals zero: ${2r^2-k^2}=0$
You get no intercepts if the discriminant is negative: ${2r^2-k^2}<0$

Solve for k to answer the question.

EB

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