How do I Find the Co-ordinates of a point of the Line AB with the circle?
The Gradient of AB is 1 and the centre of the circle is 5,4
You will have to rewrite your question in a clear manor.
What is the radius of the circle? Or is the radius just r but the center always at (5,4)?
Where does the line intersect the y axis? Or do you want a general formula for a line of gradient 1 intersecting the circle?
It does not tell us the radius of the circle. The centre is always 5,4. The formula would help
Don't know but it is on a line with gradient -1 that goes through point 5,4 I think. The radius would tell you how far up that line it is.
Where is this question from? Is it an exercise question or problem which you are trying to solve?
It seems strange that half of it is general and half of it is specific. Copy the entire question down rather than re-wording it yourself as you are not clear in your explanation of the all of the details of the problem.
x = y plus 1 where (x-5)^2 plus (y-4)^2 < r^2 .
Is that really all that's in the question?
Ok
The line joining the points A (0,5) and B (4,1) is a tangent to a circle whose centre C is at the point (5,4)
A. Equation of line AB I got y = 4x -3
B. Find the equation of the line through C which is perpendicular to AB I got y = -x +9 (not sure if it's right)
C.Find the co-ordinates of the point of contact of the line AB with the circle (This one i'm trying to work out)
In general
formula for a circle centred (a,b) with radius r is
(x-a)^2+(y-b)^2=r^2
Formula for a straight line of gradient m intersecting the y axis at point c is
y=mx+c
So
(x-a)^2+(mx+c-b)^2=r^2
x^2-2a+a^2+m^2x^2 +2mcx+c^2 -2(mx+c)+b^2=r^2
Now solve the quadratic. Plenty of stuff on line for you to learn how to do that. I'm not giving you the entire answer as you will not learn anything.
Ah ok scap my last answer. One min
A) y=mx+c
m=change in y / change in x
m = (1-5)/(4-0)
=-1
y=-x+c
sub in point A
5=c
y=-x+5
1) y = 5-x
2) y = x-1
3) set 5-x = x-1 to rearrange and get (3,2)
B)
Less explicit this time.
Gradient of a perpendicular line is
-1/m
i.e if the gradient of a line is 4 the gradient of a line perpendicular to it is -1/4. So work out the gradient in your case give the answer to part A.
You can now work out the gradient of your perpendicular line and use y = mx+c plus your information about point C=(5,4) to find the y intercept on your perpendicular line.
OkThe line joining the points A (0,5) and B (4,1) is a tangent to a circle whose centre C is at the point (5,4)
A. Equation of line AB I got y = 4x -3
B. Find the equation of the line through C which is perpendicular to AB I got y = -x +9 (not sure if it's right)C.Find the co-ordinates of the point of contact of the line AB with the circle (This one i'm trying to work out)
Blimey! That's a rather different question from what you originally asked...
A: y = 5 - x
B: y = x - 1
C: (3,2)
I can explain the answer if you need it, but the key is in the fact that from (4,1) to (5,0) involves a slope of -1 for y = mx + C. If the slope is -1, then perpendicular must have a slope of +1 and pass through (5,4).
Don't tell the OP everything or he will not learn anything.
C)
As we know that AB is tangential to the circle and our line in B) is perpendicular to AB and intersects C the intersection of these two lines will be the point you are looking for. Draw a diagram if you have not already. Solve simultaneous equations.
Ok I understand A,B it's just C now
I don't want to be told the answer just need abit of help getting there
Well now you know that for A) y = 5-x and for B) y = x-1, these lines are perpendicular. you are looking for the combination of x and y that meet, so set 5 - x = x - 1 and rearrange this expression as follows:
Add "x" to both sides
Add 1 to both sides
simplify to find x
Now substitute this value of x into either of the equations for the lines (since they must meet at this value) and you will get the value of y. I'd pick the second y = x - 1.
That's your mission, it's called solving simultaneous equations, and these aren't too hard. Follow the steps, and we can leave some harder problems to check understanding. That's what I do with my son with his homework.
"A" Level Maths with Mechanics?
What's that doing on here?