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 July 22nd, 2008, 04:59 PM #1 Newbie   Joined: Jul 2008 Posts: 10 Thanks: 0 Points of intersection of a circle. What would the points of intersection of the circle x^2 + ( y - 1 )^2 = 25 and y = - 7x + 26 be using the system of equations to solve?
 July 23rd, 2008, 02:59 AM #2 Newbie   Joined: Jun 2008 From: New Zealand Posts: 20 Thanks: 0 Re: Points of intersection of a circle. Substituting $y= -7x + 26$ into $x^2 + (y - 1)^2= 25$, we have $x^2 + (-7x + 26 - 1)^2= 25$ $x^2 + (-7x + 25)^2= 25$ $x^2 + 49x^2 - 350x + 625= 25$ $50x^2 - 350x +625= 25$ $50x^2 - 350x +600= 0$ $x^2 - 7x +12= 0$ $(x - 3)(x - 4)= 0$ So $x_1= 3$ or $x_2= 4$ When $x_1= 3$, $y_1= -7x_1 + 26 = 5$ when $x_2= 4$, $y_2= -7x_2 + 26 = -2$ Thus the two points are (3, 5) and (4, -2)
 July 23rd, 2008, 10:49 AM #3 Newbie   Joined: Jul 2008 Posts: 10 Thanks: 0 Re: Points of intersection of a circle. Thank you so much. How would i find the length of the chord in radical form for that answer?
 July 23rd, 2008, 02:14 PM #4 Newbie   Joined: Jun 2008 From: New Zealand Posts: 20 Thanks: 0 Re: Points of intersection of a circle. The length of the chord is the distance between (3, 5) and (4, -2). To find the distance between $(x_1, y_1)$ and $(x_2, y_2)$, use the following formula: $D= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$ The answer in radical form is $\sqrt{50}= 5\sqrt{2}$
 July 23rd, 2008, 04:11 PM #5 Newbie   Joined: Jul 2008 Posts: 10 Thanks: 0 Re: Points of intersection of a circle. Would i do the same kind of a thing to find the distance from the center of the circle to the chord?

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