# Lines Parallel to Diameters

• Jul 6th 2011, 04:33 AM
ughprecalc
Lines Parallel to Diameters
I'm pretty sure I understand these but still having trouble with some...
So if someone could help me?

Find and equation of the line that contains a diameter of a circle x^2+y^2+2x-4y=4,
and which is parallel to the line 3x+5y=4.

I can get as far as factoring but I think I'm doing it wrong. Help please? :)

And I'm not sure if this has anything to do with the lesson but our teacher asked us to solve it.

Let w,x,y, and z be positive real numbers. Prove that
(w+x)(x+y)(y+z)(z+w)>16wxyz

Second one isn't as important but if someone could help me understand it, I would really appreciate it. :D
• Jul 6th 2011, 04:43 AM
Plato
Re: Lines Parallel to Diameters
Quote:

Originally Posted by ughprecalc
Find and equation of the line that contains a diameter of a circle x^2+y^2+2x-4y=4,
and which is parallel to the line 3x+5y=4.

I can get as far as factoring but I think I'm doing it wrong.

I am not sure that I understand that last sentence.
So I will give you the factored form.
$\displaystyle (x+1)^2+(y-2)^2=4+1+4$
• Jul 6th 2011, 05:44 AM
HallsofIvy
Re: Lines Parallel to Diameters
A diameter must contain the center of the circle and Plato made it clear what that is. To be parallel to 3x+ 5y= 4, the line must have the same slope. Do you know how to find the equation of a line given a point and the slope?
• Jul 6th 2011, 05:51 AM
Soroban
Re: Lines Parallel to Diameters
Hello, ughprecalc!

Quote:

$\displaystyle \text{Find an equation of the line that contains a diameter of circle}$
$\displaystyle x^2+y^2+2x-4y\:=\:4\:\text{ and which is parallel to the line }3x+5y\:=\:4$

To write the equation of a line, we need a point on the line and the slope.

Our line is a diameter of that circle.
. . Hence, it passes through the center.

We have: .$\displaystyle (x^2 + 2x \qquad)+ (y^2 - 4y\qquad) \:=\:4$

Complete the square: .$\displaystyle (x^2 + 2x + 1) + (y^2 - 4y + 4) \:=\:4 + 1 + 4$

And we have: .$\displaystyle (x+1)^2 + (y-2)^2 \:=\:9$

The circle has center $\displaystyle C(\text{-}1,2)$ and radius $\displaystyle 3.$

$\displaystyle \text{The given line is: }\:3x + 5y \:=\:4 \quad\Rightarrow\quad y \:=\:\text{-}\tfrac{3}{5}x + \tfrac{4}{5}$
. . $\displaystyle \text{Its slope is: }\:m \,=\,\text{-}\tfrac{3}{5}$

$\displaystyle \text{Now write the equation of the line through }(\text{-}1,2)\text{ with slope }\text{-}\tfrac{3}{5}$

• Jul 6th 2011, 06:06 AM
Re: Lines Parallel to Diameters
if you have covered that on your syllabus.

$\displaystyle \frac{a+b}{2} \ge \sqrt{ab}$

$\displaystyle \left(\frac{w+x}{2}\right)\;\left(\frac{x+y}{2} \right)\;\left(\frac{y+z}{2}\right)\;\left(\frac{z +w}{2}\right)\ge\sqrt{wx}\sqrt{xy}\sqrt{yz}\sqrt{z w}$

$\displaystyle (w+x)(x+y)(y+z)(z+w)\ge\ 16\sqrt{w^2x^2y^2z^2}$
• Jul 6th 2011, 10:17 AM
earboth
Re: Lines Parallel to Diameters
Quote:

Originally Posted by ughprecalc
I'm pretty sure I understand these but still having trouble with some...
So if someone could help me?

Find and equation of the line that contains a diameter of a circle x^2+y^2+2x-4y=4,
and which is parallel to the line 3x+5y=4.

...

A parallel to the given line has the form

3x + 5y = c where c is a constant.

Since the line has to pass through the center of the circle the coordinates of the center must satisfy this equation. Plug in the coordinates of the center (see Plato's calculations) to determine c:

$\displaystyle 3 \cdot (-1) + 5 \cdot 2 = 7$

Therefore the equation of the parallel is: 3x + 5y = 7.
• Jul 6th 2011, 03:38 PM
ughprecalc
Re: Lines Parallel to Diameters
sorry everyone, i totally figured it out then fell asleep, but thank you, now i know the work i did WAS right, cause i did have second thoughts :D and as for the second problem, thanks, i'll just ask my teacher about that one, cause i wouldnt think of that up on my own xD
• Jul 7th 2011, 06:33 AM
Re: Lines Parallel to Diameters
We may also approach the 2nd problem using the fact that all squares are positive,
or at worst zero.

$\displaystyle \left(\sqrt{a}-\sqrt{b}\right)^2\ge\ 0$

The above is zero when a=b and positive otherwise.

Hence

$\displaystyle \left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-2\sqrt{a}\sqrt{b}+b$

$\displaystyle \Rightarrow\ a-2\sqrt{ab}+b\ge\ 0$

$\displaystyle \Rightarrow\ a+b\ge\ 2\sqrt{ab}$

Then

$\displaystyle (w+x)(x+y)(y+z)(z+w)\ge\ 2\sqrt{wx}\;2\sqrt{xy}\;2\sqrt{yz}\;2\sqrt{zw}$

and simplify