# Coordinate geometry

• Apr 1st 2008, 06:53 AM
Coach
Coordinate geometry
Dear forum members,

Find the values of b for which the straight line $\displaystyle y=-x+b$and the circle $\displaystyle x^2+y^2-2x-2y-0.25=0$ have no common points.

So the distance from the line to the centre of the circle has to be more than the radius, right?

I solved and obtained $\displaystyle b<2-\frac{3}{2}*\sqrt{2}$

$\displaystyle b<2-\frac{3}{2}*\sqrt{2}$
$\displaystyle b>2+\frac{3}{2}*\sqrt{2}$

Why? Because of the square root?

• Apr 1st 2008, 07:41 AM
wingless
When you get $\displaystyle \sqrt{ {\left ( \frac{b}{2} -1 \right )}^2 + {\left ( \frac{b}{2} -1 \right )}^2 } < \frac{3}{2}$,

Notice that $\displaystyle \sqrt{x^2}$ is NOT $\displaystyle x$, but it's $\displaystyle |x|$.

$\displaystyle \sqrt{ {\left ( \frac{b}{2} -1 \right )}^2 + {\left ( \frac{b}{2} -1 \right )}^2 } < \frac{3}{2}$

$\displaystyle \sqrt{ 2\cdot{\left ( \frac{b}{2} -1 \right )}^2 } < \frac{3}{2}$

$\displaystyle \sqrt{2}\cdot\sqrt{\cdot{\left ( \frac{b}{2} -1 \right )}^2 } < \frac{3}{2}$

$\displaystyle \sqrt{2} \left | \frac{b}{2} -1 \right | < \frac{3}{2}$

Now solving this gives the correct answer.
• Apr 1st 2008, 07:58 AM
Soroban
Hello, Coach!

Quote:

Find the values of $\displaystyle b$ for which the straight line $\displaystyle y\:=\:-x+b$
and the circle $\displaystyle x^2+y^2-2x-2y-0.25\:=\:0$ have no common points.

So the distance from the line to the centre of the circle
has to be more than the radius, right? . . . . Right!

As you found, the center is $\displaystyle (1,1)$ and the radius is $\displaystyle \frac{3}{2}$

I used this distance formula . . .

The distance from a point $\displaystyle (x_1,y_1)$ to the line $\displaystyle ax + by + c \:=\:0$

. . is given by: .$\displaystyle d \;=\;\frac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}$ . .
Note the absolute value.

We have point $\displaystyle (1,1)$ and line $\displaystyle x + y - b \:=\:0$

The distance is: .$\displaystyle d \;=\;\frac{|1 + 1 - b|}{\sqrt{1^2+1^2}}$ . . . which is to be greater than $\displaystyle \frac{3}{2}$

So we have: .$\displaystyle \frac{|2-b|}{\sqrt{2}} \:>\:\frac{3}{2}\quad\Rightarrow\quad |2 - b| \:> \:\frac{3\sqrt{2}}{2}$

And we have two inequalities to solve . . .

. . $\displaystyle 2 - b \:>\:\frac{3\sqrt{2}}{2}\quad\Rightarrow\quad \text{-}b \:> \:\text{-}2 + \frac{3\sqrt{2}}{2} \quad\Rightarrow\quad\boxed{ b \:< \:2 - \frac{3\sqrt{2}}{2}}$

. . $\displaystyle 2-b \:< \:\text{-}\frac{3\sqrt{2}}{2}\quad\Rightarrow\quad \text{-}b \:< \:\text{-}2 - \frac{3\sqrt{2}}{2} \quad\Rightarrow\quad \boxed{b \:> \:2 + \frac{3\sqrt{2}}{2}}$