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    I have worked out the centre of the circle (3,2) and the radius is 2. I think I have to use b^2 -4ac . Just don't know what appoach to use.
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    (Original post by sunnytunny92)
    I have worked out the centre of the circle (3,2) and the radius is 2. I think I have to use b^2 -4ac . Just don't know what appoach to use.
    Centre looks like (3,-2), but it's rather difficult to see from the picture. Can you post a clearer one. It's question 10 I presume.
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    (Original post by ghostwalker)
    Centre looks like (3,-2), but it's rather difficult to see from the picture. Can you post a clearer one. It's question 10 I presume.
    sorry its q11. yh and sorry again for picture clarity, have had to use my laptop camera. ok ive just worked out part (a) and got the correct solution.

    for (a) just subbed in y=mx in the circle equation, then subbed in the values of a,b and c into b^2 -4ac=0, then factored it and got there in the end. just working out (b) and (c) thanks
 
 
 
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