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# how do you get the equation of a circle if only given the line that goes through the centre and that x axis and y axis are tangents?

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x + 2y - 6 = 0

Feb 3, 2015

#2
+21860
+10

x + 2y - 6 = 0

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

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Feb 3, 2015

#1
+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
+21860
+10
$$x + 2y - 6 = 0\\ 2y=-x+6\\ \boxed{y=f(x)=-\frac{1}{2}x+3}\\\\ \small{\text{ r = radius of the circle }}\\ \small{\text{  x=f(x)=r  }}\\ \small{\text{  x=-\frac{1}{2}x+3  }}\\ \small{\text{\frac{1}{2}x+x=3 }}\\ \small{\text{\frac{3}{2}x=3 }}\\ \small{\text{x=\frac{2}{3}*3=2}}\\ \small{\text{r=2. The center of the circle is (x_m,\ y_m) =\ ( 2,\ 2) }}\\ \small{\text{The equation of the circle is: }}\\ \small{\text{  (x-x_m)^2+(y-y_m)^2=r^2 \qquad (x-2)^2+(y-2)^2=2^2  }}$$