+0  
 

Best Answer 

 #2
avatar+21860 
+10

x + 2y - 6 = 0

$$x + 2y - 6 = 0\\
2y=-x+6\\
\boxed{y=f(x)=-\frac{1}{2}x+3}\\\\
\small{\text{
r = radius of the circle
}}\\
\small{\text{
$
x=f(x)=r
$
}}\\
\small{\text{
$
x=-\frac{1}{2}x+3
$
}}\\
\small{\text{$\frac{1}{2}x+x=3 $}}\\
\small{\text{$\frac{3}{2}x=3 $}}\\
\small{\text{$x=\frac{2}{3}*3=2$}}\\
\small{\text{$r=2$. The center of the circle is $(x_m,\ y_m) =\ ( 2,\ 2)$
}}\\
\small{\text{The equation of the circle is:
}}\\
\small{\text{
$
(x-x_m)^2+(y-y_m)^2=r^2 \qquad (x-2)^2+(y-2)^2=2^2
$
}}$$

.
 Feb 3, 2015
 #1
avatar+98173 
+10

We're looking for a point on the line that is an equal distance to both axis..........thus, we're looking for a point such that x = y

So, we have

x + 2y - 6 = 0      let y = x

x + 2x - 6  = 0

3x - 6  = 0

3x = 6

x = 2

And by the same rationale, y= 2

Thus, our center is (2,2)   and the radius = 2

Here's a graph...

GRAPH

 

 Feb 3, 2015
 #2
avatar+21860 
+10
Best Answer

x + 2y - 6 = 0

$$x + 2y - 6 = 0\\
2y=-x+6\\
\boxed{y=f(x)=-\frac{1}{2}x+3}\\\\
\small{\text{
r = radius of the circle
}}\\
\small{\text{
$
x=f(x)=r
$
}}\\
\small{\text{
$
x=-\frac{1}{2}x+3
$
}}\\
\small{\text{$\frac{1}{2}x+x=3 $}}\\
\small{\text{$\frac{3}{2}x=3 $}}\\
\small{\text{$x=\frac{2}{3}*3=2$}}\\
\small{\text{$r=2$. The center of the circle is $(x_m,\ y_m) =\ ( 2,\ 2)$
}}\\
\small{\text{The equation of the circle is:
}}\\
\small{\text{
$
(x-x_m)^2+(y-y_m)^2=r^2 \qquad (x-2)^2+(y-2)^2=2^2
$
}}$$

heureka Feb 3, 2015

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