SOLUTION: Hi there,really stumped by this question. For a circle with the equation x^2 + y^2 - 2x - 14y + 25 = 0, show that if the line y = mx + c intersects the circle at two points, then (

Algebra ->  Circles -> SOLUTION: Hi there,really stumped by this question. For a circle with the equation x^2 + y^2 - 2x - 14y + 25 = 0, show that if the line y = mx + c intersects the circle at two points, then (      Log On


   

Question 1087984: Hi there,really stumped by this question. For a circle with the equation x^2 + y^2 - 2x - 14y + 25 = 0, show that if the line y = mx + c intersects the circle at two points, then (1 + 7m)^2 - 25(1 + m^2) > 0.
I can work out the centre of the circle to be (1,7) and the radius as 5, but I cannot see that helps.
Paul
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
The idea on how to solve this problem is THIS: 

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

    %28mx%2Bc%29%5E2+-+2x+-+14%28mx%2Bc%29+%2B+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.

I completed my tutor's instructions.

You implement this guiding idea.