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    11 A circle has equation (x − 3)^2 + (y + 2)^2 = 25.

    (i) State the coordinates of the centre of this circle and its radius. [2]

    (ii) Verify that the point A with coordinates (6, −6) lies on this circle. Show also that the point B on the circle for which AB is a diameter has coordinates (0, 2). [3]

    (iii) Find the equation of the tangent to the circle at A. [4]

    (iv) A second circle touches the original circle at A. Its radius is 10 and its centre is at C, where BAC is a straight line. Find the coordinates of C and hence write down the equation of this second circle. [3]

    I've done all parts apart from iv). I'd be very grateful if someone could explain how to do part iv) in detail.

    Thanks very much.
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    i) Looking at the circle equation, you can work out the coordinates. I.e X-3=0, therefore x co-ordinate is..... Y+2=0, therefore y co-ordinate is....

    ii) Put (6,-6) into the equation of the circle.

    iii) Using part (i) and (6,-6), work out the gradient of the tangent and then the equation using y-y1=m(x-x1)

    iv) Draw a diagram, it really is that easy. Basically, it is (6, -6), 10 across, so the x-coordinate will vary. You know it (the equation) =100 as radius is 10. Have fun.
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    Iv) isn't that simple

    (Original post by Newcastle456)
    i) Looking at the circle equation, you can work out the coordinates. I.e X-3=0, therefore x co-ordinate is..... Y+2=0, therefore y co-ordinate is....

    ii) Put (6,-6) into the equation of the circle.

    iii) Using part (i) and (6,-6), work out the gradient of the tangent and then the equation using y-y1=m(x-x1)

    iv) Draw a diagram, it really is that easy. Basically, it is (6, -6), 10 across, so the x-coordinate will vary. You know it (the equation) =100 as radius is 10. Have fun.
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    (Original post by iramchaudhry94)
    11 A circle has equation (x − 3)^2 + (y + 2)^2 = 25.

    (i) State the coordinates of the centre of this circle and its radius. [2]

    (ii) Verify that the point A with coordinates (6, −6) lies on this circle. Show also that the point B on the circle for which AB is a diameter has coordinates (0, 2). [3]

    (iii) Find the equation of the tangent to the circle at A. [4]

    (iv) A second circle touches the original circle at A. Its radius is 10 and its centre is at C, where BAC is a straight line. Find the coordinates of C and hence write down the equation of this second circle. [3]

    I've done all parts apart from iv). I'd be very grateful if someone could explain how to do part iv) in detail.

    Thanks very much.
    B, A, and C are collinear, so gradient of AC = gradient of BA = (2 - -6)/(0-6) = -8/6 = -4/3, so the vector AC is (-4k, 3k) for some k, and thus the length of AC is 5k (using the 3,4,5 Pythagorean triple). But we know C is the centre of the second circle and A is on the second circle, so AC is a radius of this circle, and hence the length of AC is 10. Thus 5k = 10, so k = 2, giving the vector AC as (-8, 6), and thus C = (6-8, -6+6) = (-2,0).
 
 
 
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