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Thread: Equation of the Line & Two Circles

  1. #1
    Senior Member harpazo's Avatar
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    Equation of the Line & Two Circles

    Find an equation of the line containing the centers of the two circles given below.

    x^2 + y^2 - 4x + 6y + 4 = 0 and

    x^2 + y^2 + 6x + 4y + 9n= 0.

    Seeking the first-two steps.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Equation of the Line & Two Circles

    1.) Put both circles in the form:

    $\displaystyle (x-h)^2+(y-k)^2=r^2$

    2.) Use the formula:

    $\displaystyle y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1$
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  3. #3
    Senior Member harpazo's Avatar
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    Re: Equation of the Line & Two Circles

    Quote Originally Posted by MarkFL View Post
    1.) Put both circles in the form:

    $\displaystyle (x-h)^2+(y-k)^2=r^2$

    2.) Use the formula:

    $\displaystyle y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1$
    I understand step 1. Are the values for x_1 & y_1 needed for step 2 found in your first step?
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    Re: Equation of the Line & Two Circles

    Quote Originally Posted by harpazo View Post
    I understand step 1. Are the values for x_1 & y_1 needed for step 2 found in your first step?
    The second formula MarkFL gave you is used to find the equation of the line through the two points (x1, y1) and (x2, y2).

    You are required to find the equation of the line through the two circle centres - so what are those two points?
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  5. #5
    Senior Member harpazo's Avatar
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    Re: Equation of the Line & Two Circles

    Quote Originally Posted by harpazo View Post
    I understand step 1. Are the values for x_1 & y_1 needed for step 2 found in your first step?
    The first equation is (x - 2)^2 + (y + 3)^2 = 25. The center (h, k) is (2, -3).

    The second equation is (x + 3)^2 + (y + 2)^2 = 16. The center (h, k) is (-3, -2).

    Let x_1 = 2, y_1 = -3, x_2 = -3 and y_2 = -2.

    We now plug and chug.

    y = [(-2 -(-3))/(-3 -2)](x - 2) - 3

    y = [(-2 + 3)/-5](x - 2) - 3

    y = (-1/5)(x - 2) - 3

    y = [-(x - 2)/5] - 3

    Correct?
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    Senior Member harpazo's Avatar
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    Re: Equation of the Line & Two Circles

    Quote Originally Posted by Debsta View Post
    The second formula MarkFL gave you is used to find the equation of the line through the two points (x1, y1) and (x2, y2).

    You are required to find the equation of the line through the two circle centres - so what are those two points?
    Read my reply to MarkFL.
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    Re: Equation of the Line & Two Circles

    Quote Originally Posted by harpazo View Post
    The first equation is (x - 2)^2 + (y + 3)^2 = 25. this should be 9.The center (h, k) is (2, -3). Centre is correct.

    The second equation is (x + 3)^2 + (y + 2)^2 = 16 this should be 4. The center (h, k) is (-3, -2). Centre is correct.

    Let x_1 = 2, y_1 = -3, x_2 = -3 and y_2 = -2.

    We now plug and chug.

    y = [(-2 -(-3))/(-3 -2)](x - 2) - 3

    y = [(-2 + 3)/-5](x - 2) - 3

    y = (-1/5)(x - 2) - 3

    y = [-(x - 2)/5] - 3

    Correct?
    Final equation is correct. You are not getting the centre/radius from of the circle correct though.


    Equations of lines are usually written in:

    gradient-intercept form: y=mx + c , in this case $\displaystyle y=\frac{-1}{5}x - \frac{13}{5}$

    or

    standard form: ax+ by +c =0 , in this case x + 5y + 13 =0
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  8. #8
    Senior Member harpazo's Avatar
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    Re: Equation of the Line & Two Circles

    Quote Originally Posted by Debsta View Post
    Final equation is correct. You are not getting the centre/radius from of the circle correct though.


    Equations of lines are usually written in:

    gradient-intercept form: y=mx + c , in this case $\displaystyle y=\frac{-1}{5}x - \frac{13}{5}$

    or

    standard form: ax+ by +c =0 , in this case x + 5y + 13 =0
    Thank you for pointing out my error(s) in red. This is perfect for my notes. I will practice more and more as I go through the Michael Sullivan textbook.
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