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The lines 3x+y=1 and 5x-y=15 intersect at the center of circle O. If the circle is tangent to the y-axis, find the equation of the circle.

 May 27, 2015

Best Answer 

 #2
avatar+20850 
+13

The lines 3x+y=1 and 5x-y=15 intersect at the center of circle O. If the circle is tangent to the y-axis, find the equation of the circle.

$$\\
\small{\text{
(1. Line):
$
\begin{array}{rcl}
\\\\\\
3x + y &=& 1 \\
y &=& \underbrace{-3}_{m_1}x + \underbrace{1}_{b_1} \\
\end{array}
$}} \\
\small{\text{
(2. Line):
$
\begin{array}{rcl}
\\\\\\
5x - y &=& 15 \\
y &=& \underbrace{5}_{m_2}x \underbrace{-15}_{b_2} \\
\end{array}
$}}\\\\
\small{\text{
Intersection:
$
\begin{array}{lcl}
\boxed{
~x_{\rm{intersection}}=- \dfrac{\Delta b}{ \Delta m} = - \dfrac{b_1-b_2}{m_1-m_2} ~
}
\\\\\\
x_{\rm{intersection}}=- \dfrac{1-(-15)}{-3-5} = -\dfrac{16}{-8} = \dfrac{16}{8}= 2\\
y_{\rm{intersection}} = -3x+1 \\
y_{\rm{intersection}} = -3\cdot 2+1\\
y_{\rm{intersection}} = -6+1\\
y_{\rm{intersection}} = -5
\end{array}
$}}\\\\$$

$$\\\small{\text{
Circle center $(x_c,y_c):
\begin{array}{lcl}
\\\\\\
x_c=x_{\rm{intersection}}=2\\
y_c=y_{\rm{intersection}} = -5
\end{array}
$}}\\\\
\small{\text{
Circle radius $r:
\begin{array}{lcl}
\\\\
r= x_c=x_{\rm{intersection}}=2\\
\end{array}
$}}\\\\
\small{\text{
Circle formula:
$
\begin{array}{lcl}
\\\\
(x-x_c)^2+(y-y_c)^2=r^2\\
(x-2)^2+(y+5)^2=2^2=4\\
\end{array}
$}}$$

.
 May 28, 2015
 #1
avatar+94611 
+10

3x+y=1 and 5x-y=15

 

Using the first equation, y = 1 - 3x    ....and substituting this into the second, we have

 

5x - (1 - 3x) = 15

 

5x -1 + 3x = 15

 

8x - 1  = 15

 

8x = 16

 

x = 2     and y = (1 - 3x) = (1 - 3(2))  = (1 - 6)  = -5

 

So the solution point is (2, -5)

 

And the equation of the circle would be.....

 

(x - 2)^2 + (y + 5)^2  = 4

 

Here's a graph.....https://www.desmos.com/calculator/pt2wwhqn4u

 

 

 

 May 27, 2015
 #2
avatar+20850 
+13
Best Answer

The lines 3x+y=1 and 5x-y=15 intersect at the center of circle O. If the circle is tangent to the y-axis, find the equation of the circle.

$$\\
\small{\text{
(1. Line):
$
\begin{array}{rcl}
\\\\\\
3x + y &=& 1 \\
y &=& \underbrace{-3}_{m_1}x + \underbrace{1}_{b_1} \\
\end{array}
$}} \\
\small{\text{
(2. Line):
$
\begin{array}{rcl}
\\\\\\
5x - y &=& 15 \\
y &=& \underbrace{5}_{m_2}x \underbrace{-15}_{b_2} \\
\end{array}
$}}\\\\
\small{\text{
Intersection:
$
\begin{array}{lcl}
\boxed{
~x_{\rm{intersection}}=- \dfrac{\Delta b}{ \Delta m} = - \dfrac{b_1-b_2}{m_1-m_2} ~
}
\\\\\\
x_{\rm{intersection}}=- \dfrac{1-(-15)}{-3-5} = -\dfrac{16}{-8} = \dfrac{16}{8}= 2\\
y_{\rm{intersection}} = -3x+1 \\
y_{\rm{intersection}} = -3\cdot 2+1\\
y_{\rm{intersection}} = -6+1\\
y_{\rm{intersection}} = -5
\end{array}
$}}\\\\$$

$$\\\small{\text{
Circle center $(x_c,y_c):
\begin{array}{lcl}
\\\\\\
x_c=x_{\rm{intersection}}=2\\
y_c=y_{\rm{intersection}} = -5
\end{array}
$}}\\\\
\small{\text{
Circle radius $r:
\begin{array}{lcl}
\\\\
r= x_c=x_{\rm{intersection}}=2\\
\end{array}
$}}\\\\
\small{\text{
Circle formula:
$
\begin{array}{lcl}
\\\\
(x-x_c)^2+(y-y_c)^2=r^2\\
(x-2)^2+(y+5)^2=2^2=4\\
\end{array}
$}}$$

heureka May 28, 2015
 #3
avatar+4682 
0

The Latex is impeccable!

 May 28, 2015
 #4
avatar+95361 
0

Yes, Heureka is the master of LaTex.  His maths is not half bad either LOL  

Thanks Chris and Heureka  

 May 28, 2015

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