circle intersection, find points

• December 12th 2008, 09:40 AM
booper563
circle intersection, find points
Find the points of intersetion of the circle x^2+(y-1)^2=25 and y=-7x+26 by solving a system of equations.

So far I got x^2 +(-7x+26-1)^2=25
x^2+(7x^2)+26x-1=25

now what do I do am I doing it right?

B) Find the length of the chord determined by the poibnts of intersection in question a. Express your answer in simpliest radical form.

C) Determine the distance of the chord from the centre of the circle

Like i said before this math is REALLY confusing to me, so details and step by step instructions would be amazing help.
thanks guys
• December 12th 2008, 09:50 AM
chabmgph
Hi booper563,
Quote:

Find the points of intersetion of the circle x^2+(y-1)^2=25 and y=-7x+26 by solving a system of equations.

So far I got $x^2 +(-7x+26-1)^2=25$
$x^2+(7x^2)+26x-1=25$

now what do I do am I doing it right?
The first line looks okay, but the second line isn't.
Note that $(a+b)^2 = a^2+2ab+b^2$
• December 12th 2008, 09:57 AM
moolimanj
for the points of intersection you could use the method you used.

when both equations are combined you should get:

50x^2-350x+600=0

this factorises to: 50(x-4)(x-3)=0

hence x=4 and x=3 i.e. there are two points as you would expect (unless the circles are touching in which case there will be one).

Then substitute values of 3 and 4 into equations to get y and coordinates at each end.

The rest of the question should be relatively straight forward from here on.
• December 12th 2008, 10:57 AM
Soroban
Hello, booper563!

Quote:

A) Find the points of intersection of the circle $x^2+(y-1)^2\:=\:25\,\text{ and }\,y\:=\:-7x+26$

We have: . $x^2 + (-7x+26-1)^2 \:=\:25 \quad\Rightarrow\quad x^2 + (-7x + 25)^2 \:=\:25$

Simplify: . $x^2 + 49x^2 - 350x + 625 \:=\:25 \quad\Rightarrow\quad 50x^2 - 350x + 600\:=\:0$

Divide by 50: . $x^2 - 7x + 12 \:=\:0 \quad\Rightarrow\quad (x-3)(x-4) \:=\:0$

And we have: . $\begin{Bmatrix}x \,=\, 3 & \Rightarrow & y \,=\, 5 \\ x \,=\, 4 & \Rightarrow & y \,=\, \text{-}2 \end{Bmatrix}$

. . The intersections are: . $A(3,5),\;B(4,\text{-}2)$

Quote:

B) Find the length of the chord determined by the points of intersection in A.

Use the Distance Formula on points $A$ and $B.$

. . $\overline{AB} \;=\;\sqrt{(4-3)^2 + (\text{-}2-5)^2} \;=\;\sqrt{1 + 49} \;=\;\sqrt{50} \;=\;5\sqrt{2}$

Quote:

C) Determine the distance of the chord from the centre of the circle.
A handy fact: the radius to the midpoint of a chord is perpendicular to the chord.

The midpoint of AB is: . $M\left(\tfrac{7}{2},\,\tfrac{3}{2}\right)$

We want the distance from the center $C(0,1)$ to $M.$

. . $\overline{CM} \;=\;\sqrt{\left(\tfrac{7}{2}\right)^2 + \left(\tfrac{3}{2}-1\right)^2} \;=\; \sqrt{\tfrac{49}{4} + \tfrac{1}{4}} \;=\;\sqrt{\tfrac{50}{4}} \;=\; \frac{5\sqrt{2}}{2}$