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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.

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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

#2
+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.

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

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

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May 28, 2015

#1
+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
+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.

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

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

heureka May 28, 2015
#3
+4682
0

The Latex is impeccable!

May 28, 2015
#4
+95361
0

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

Thanks Chris and Heureka

May 28, 2015