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    A circle has equation (x+3)^2 + (y-4)^2 =25
    (1) state the centre and radius of the circle.
    (2) find the co-ordinates of the point of intersection of the line y= 2x+5 with the circle.
    (3) find the equation of the tangent to the circle at the point(1,1)
    (4) Find the point of the circle which has the point (1,1) at the other end of its diameter.
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    (Original post by JOHN EK)
    A circle has equation (x+3)^2 + (y-4)^2 =25
    (1) state the centre and radius of the circle.
    (2) find the co-ordinates of the point of intersection of the line y= 2x+5 with the circle.
    (3) find the equation of the tangent to the circle at the point(1,1)
    (4) Find the point of the circle which has the point (1,1) at the other end of its diameter.
    We aren't going to do this Q for you. Have you read the guidelines?

    Post you attempt/thoughts on the part you want help with.
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    Note a circle has general form (x-a)^2 + (y-b)^2 = r^2.

    Do you see how we can find the centre and radius now?

    2) Equate and solve.

    3) Either use implicit differentiation and evaluate gradient of tangent that way or just do -1/gradient of radius.

    4) Use the fact that that it is a diameter.
 
 
 
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