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    can someone help me to work andd understand parts 3 and 4 please,



    on part 3, i dont understand the formula shown on the mark scheme
    and on that part, i even drew out the points and the circle on squared paper and the point was outside, not inside as it says in the mark scheme...
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    Part 3 simply uses the distance from (-3,9) to the centre of the circle to determine whether or not it is within the radius.

    For part 4, do you know how to find the gradient of a line that is perpendicular to another?
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    (Original post by Hopple)
    Part 3 simply uses the distance from (-3,9) to the centre of the circle to determine whether or not it is within the radius.

    For part 4, do you know how to find the gradient of a line that is perpendicular to another?
    its just for part three, it says on the mark scheme that its inside but i drew it out and it was outside...
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    I agree with the mark scheme

    The circle contains all points within 10 units of the centre, so by showing that (-3,9) is closer than 10 units away from the centre shows that it is within the circle.
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    (Original post by Hopple)
    I agree with the mark scheme

    The circle contains all points within 10 units of the centre, so by showing that (-3,9) is closer than 10 units away from the centre shows that it is within the circle.
    hmm yeah, i fully understand part 3 now but part 4, i dont get why i have to get the gradient of the radius? then find the normal of that when it says tangent
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    http://www.youtube.com/watch?v=WVnwEMK66zE
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    (Original post by cooldudeman)
    hmm yeah, i fully understand part 3 now but part 4, i dont get why i have to get the gradient of the radius? then find the normal of that when it says tangent
    Gradient of the radius is probably a confusing term for it to use. What it means is that you should obtain the gradient of the line running from the centre of the circle to the point (8,9). The tangent to any point on a circle's circumference is 90 degrees (i.e. perpendicular) to the line running from the centre. So -1 divided by the gradient of the line running from the centre is the gradient of the tangent.
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    (Original post by Sasukekun)
    Gradient of the radius is probably a confusing term for it to use. What it means is that you should obtain the gradient of the line running from the centre of the circle to the point (8,9). The tangent to any point on a circle's circumference is 90 degrees (i.e. perpendicular) to the line running from the centre. So -1 divided by the gradient of the line running from the centre is the gradient of the tangent.
    oh okok i understand what you're saying but just for the record, what would the equation of the normal be at point (8,9).
    would it just be the gradient of the line running from the centre to the point and this would be added in the equation, y-y1=gradient(x-x1)
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    (Original post by cooldudeman)
    oh okok i understand what you're saying but just for the record, what would the equation of the normal be at point (8,9).
    would it just be the gradient of the line running from the centre to the point and this would be added in the equation, y-y1=gradient(x-x1)
    Hmm, well I don't want to give the answer away. But the equation of the line from the centre of the circle to (8,9) is as you stated  y-y_1 = grad(x-x_1) Where x1 and y1 are points 8 and 9 respectively. The gradient of the normal to this line (i.e tangent) is minus one (-1) divided by the gradient of the line from the centre of the circle. So it’s equation is  y-y_1 = \frac {-1}{Grad} * (x-x_1)
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    (Original post by Sasukekun)
    Hmm, well I don't want to give the answer away. But the equation of the line from the centre of the circle to (8,9) is as you stated  y-y_1 = grad(x-x_1) Where x1 and y1 are points 8 and 9 respectively. The gradient of the normal to this line (i.e tangent) is minus one (-1) divided by the gradient of the line from the centre of the circle. So it’s equation is  y-y_1 = \frac {-1}{Grad} * (x-x_1)
    ok for the equation for the tangent of (8,9), i got y=-3/4+15. this is obviously right, as the mark scheme says. and the normal of that point would be
    y=4/3x -5/3

    am i right? if i am right, then that means that circles are very differnet to just normal lines and curves...
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    (Original post by cooldudeman)
    ok for the equation for the tangent of (8,9), i got y=-3/4+15. this is obviously right, as the mark scheme says. and the normal of that point would be
    y=4/3x -5/3

    am i right? if i am right, then that means that circles are very differnet to just normal lines and curves...

    Yes you're right, I imagine, as the mark scheme mirrors your answer. I'm not really sure what you mean by 'different'. Aye, they are different in that circles are circles and lines are lines, but what do you mean?

    Here's a picture just in case;

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