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# C1- AS Maths Circle Question Help watch

1. 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.
2. Bump
3. 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.
4. 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.
5. (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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