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.Com
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)   (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

     = 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.



RELATED QUESTIONS

Consider the circle x^2 + y^2 - 2x -14y +25=0 Show that, if the line y=mx intersects the (answered by ikleyn)
Hi Guys, really struggling with this question. Given the Cartesian equation of a... (answered by ewatrrr)
What is the center of the circle given by the equation x^2 + y^2 - 10x + 14y+ 10 =0 (answered by MathLover1)
IT'S FOR MY PRE-CALCULUS PLEASE HELP ME! 1. Find the equation of the circle... (answered by ikleyn)
I really need help with this 4 part equation.I have looked at it and just stumped could... (answered by josmiceli)
I have an equation for a circle (x+3)^2+(y-1)^2=25 where the centre point is 3,1 and th... (answered by longjonsilver)
Determine if the circles defined by each pair of equations intersect a)x^2 + y^2 - 10x + (answered by Alan3354)
Hi! I'm really having a difficulty answering this question so you might help me with... (answered by Theo)
Hello I'm really stuck on this question and would be very grateful if you could help... (answered by jim_thompson5910)