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...
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.
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
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.
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)
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−y1=grad(x−x1) 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−y1=Grad−1∗(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−y1=grad(x−x1) 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−y1=Grad−1∗(x−x1)
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...
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?