Question 1087984
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The idea on how to solve this problem is THIS:


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1.  Substitute y = mx +c into the second degree equation, replacing y. You will get

    {{{(mx+c)^2 - 2x - 14(mx+c) + 25}}} = 0.


2.  Simplify the last equation and reduce it to the standard form of a quadratic equation

    ax^2 + bx + c = 0.


3.  The fact that  "the line y = mx + c intersects the circle at two points"  means that this quadratic equation has 
    two different real solutions.


4.  This, in turn, means that the discriminant of the quadratic equation is POSITIVE: d = b^2 - 4ac > 0.


5.  This inequality is exactly what you need to prove.
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I completed my tutor's instructions.


You implement this guiding idea.