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Circles (Coordinate Geometry)

 
 
Skyshr
 
Reply Wed 7 Dec, 2005 04:54 am
Please teach me to do this question

I have a circle C: x^2 + y^2 - 2y = 0
and a circle K which touches C and the X-axis.
Let A(s,t) be the centre of K, how can I show that 4t = s^2 ?

And then how can I find the eqations of circles which ...
1. pass through the point (6/5,2/5), and ...
2. touch the x-axis and C in quadrant I?
There are 2 answers but I can't find any of them Sad

My learnings: For a circle with equation x^2 + y^2 + Dx + Ey + F = 0
the centre is (-D/2, -E/2)
the radius is [(D^2 + E^2 - 4F)^(1/2)] * 0.5
How can I use these formulae to do the question?
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engineer
 
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Reply Wed 7 Dec, 2005 11:09 am
There are a couple of assumptions I've made based on what you've posted.

1) The circle A touches circle C at only one point
2) The circle A touches the X axis at only one point

From what you've learned in class, the center of C is (0,1) and the radius of C is 1. If circle A has a center of (s,t) and touches the X axis at only one point, the radius of A is t. (If you don't see that, draw a picture and it will be clear.) This means that the distance between the center of A and the center of C is the sum of their radii or (t+1). You can use the formula for distance between two points to see that D^2 = (s-0)^2 + (t-1)^2 which is also (t+1)^2. Solve and you get that 4t=s^2.

To find circles that include (1.2, .4) you need to construct the equation for circle A. This form of the circle equation is easier to use when you know the center. (x-xc)^2 + (y-yc)^2 = R^2

You know that xc=s=2 t^.5, yc=t and R=t from your work above. x=1.2 and y=.4 is given. Solve the resulting quadratic and you will have your two circles.
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