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 May 19th, 2013, 06:02 AM #1 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Circles and Chords I can't spot where I've gone on this one. Can anyone help me out? Many thanks. Q. Find the equation of the circle with centre in the 1st quadrant, if it touches the y-axis at the point (0, 2) & makes a chord of length 3 units on the x-axis. Attempt: If the circle touches the y-axis at (0, 2) then centre c = (x, 2), Perpendicular distance of c to x-axis = 2, Let |ab| = 1/2 length of chord on x-axis = 3/2, |ac| = $(\frac{3}{2})^2+2^2=\frac{25}{4}$, Thus, from y-axis, $c\,=(\frac{25}{4},\,2)$, Equation of circle: $(x-\frac{25}{4})^2+(y-2)^2=(\frac{25}{4})^2\rightarrow x^2+y^2-\frac{25x}{2}-4y+4\,{=0$ Ans. (From text book): $x^2+y^2-5x-4y+4\,{=0$
 May 19th, 2013, 06:31 AM #2 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Re: Circles and Chords I see where I went wrong. Radius should have been 5/2. Apologies.
May 19th, 2013, 06:47 AM   #3
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Re: Circles and Chords

Quote:
 Originally Posted by bilano99 I see where I went wrong. Radius should have been 5/2. Apologies.
[color=#000000]Since you posted a problem, why don't you give a detailed proof? Others may be benefited by studying the answer. This is not a "one-way" forum![/color]

May 19th, 2013, 07:49 AM   #4
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 Originally Posted by bilano99 I can't spot where . . . |ac| = $(\frac{3}{2})^2+2^2=\frac{25}{4}$, . . .
That should have been |ac|² = etc., so that |ac| = 5/2.

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