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 October 28th, 2011, 06:39 AM #1 Newbie   Joined: Oct 2011 Posts: 2 Thanks: 0 Circle Geometry I have 3 questions which I'm unsure how to do (or have some idea but not entirely sure of all parts) on circle geometry. Help on any of these questions would be wonderful. 6) A circle passes through the points Q(0,3) and R (0,9) and touches the x axis. Work out two possible equations. Now for this I know that you need to first find the co-ordinate for where the circle touches the x axis and I will most likely know what to do from there. But I'm not sure how to go about doing this :/ 7) i) Show that the line y= 4-x is a tangent to the circle x^2 + y^2 = 8 ii) Show that the line 4y= 3x-25 is a tangent to the circle x^2+y^2 = 25 I know what a tangent is, I just don't know how you go about showing it? Find the coordinates of the point where the following lines and parabolas intersect. i) y = 3x+ 1 and y = x^2 - 4x +7 ii) y = x -2 and y = x^2 + 2x-8 I have never heard of the word parabola in my life!
 October 28th, 2011, 07:54 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,635 Thanks: 2080 6) Let S(x, y) be the centre of the circle, then it touches the x-axis at (x, 0). Can you see why y = 6? How far is S from Q and R? 7) Consider where the line and circle intersect. For a tangent, there is exactly one such point. 8) This requires solving simultaneous equations. I recommend you eliminate y and then solve for x.
 October 28th, 2011, 08:26 AM #3 Newbie   Joined: Oct 2011 Posts: 2 Thanks: 0 Re: Circle Geometry 6) Yes I think I understand why y is 6, but how do I solve for x? 7) And how would I go about this? 8) That seems simple enough, I can do that, thank you
 October 28th, 2011, 04:08 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,635 Thanks: 2080 6) You didn't answer my second question. 7) Where the graphs intersect, the coordinates satisfy both equations, so you can proceed as described for question 8, and hence find out how many solutions there are.
 October 30th, 2011, 12:49 AM #5 Senior Member   Joined: Oct 2011 From: India ???? Posts: 224 Thanks: 0 Re: Circle Geometry Q 6) Let the equation of the circle be $(x-\alpha)^2+(y-\beta)^2=r^2$. Since (0,3) and(0,9) lie on the circle: $(-\alpha)^2+(3-\beta)^2=r^2$ and $(-\alpha)^2+(9-\beta)^2=r^2$ Subtracting: $(-\alpha)^2+(9-\beta)^2=r^2$ $(-\alpha)^2+(3-\beta)^2=r^2$ ___________________________ $(9-\beta)^2-(3-\beta)^2=0$ $\Rightarrow (12-2\beta)(6)=0 \\ \Rightarrow \beta=6$ Since x-axis touches the circle: $r=\beta=6$ In the equation of circle: $(x-\alpha)^2+(y-\beta)^2=r^2 \\ \Rightarrow (-\alpha)^2+(3-6)^2=6^2 \\ \Rightarrow \alpha=\pm 3\sqrt{3}$ Therefore the centers of the circles are $(3\sqrt{3},6)$ and $(-3\sqrt{3},6)$ and the $r=6$. Q7) The line $y=4-x$ is tangent to the curve $x^2+y^2=8$ if it touches it at only 1 point. Substituting y in equation of the curve: $x^2+(4-x)^2=8 \\ \Rightarrow x^2+16+x^2-8x=8 \\ \Rightarrow 2x^2-8x+8=0 \\ \Rightarrow x^2-4x+4=0 \\ \Rightarrow x=2$ Only 1 solution exists. The line is tangent to the curve. Q(i) $y=3x+1$ and $y=x^2-4x+7$. Solving these two by substitution. $3x+1=x^2-4x+7 \\ \Rightarrow x^2-7x+6=0 \\ \Rightarrow (x-6)(x-1)=0 \\ \Rightarrow x=6,1$ Therefore $y=19,4$. So the line intersects the parabola at $(6,19)$ and $(1,4)$. (ii) Do the same as in (i).
October 30th, 2011, 04:55 AM   #6
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Quote:
 Originally Posted by Becky-RanRan 6) I think I understand why y is 6, but how do I solve for x?
It's 6 because (x, y) lies on the perpendicular bisector of QR. The circle touches the x-axis, so its radius is 6.
As the distance of (x, 6) from Q(0, 3) is 6, x² + 9 = 36, and so x = ±3?3.
Hence the equations of the possible circles are (x ± 3?3)² + (y - 6)² = 36.

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# show that the line 4y=3x-25 is a tnagent to the line x^2 y^2=25

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