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# Can someone help me answer this questions: maths - coordinate geometry watch

1. Can you help me understand the steps to finding the answers to these type of questions so that I know how to tackle future questions like these:

1) Find k so that point (3, root27) lies on the circle x^2+y^2=k^2. If P, Q and R lie on the circle, and triangle PQR is equilateral, write down the coordinates of the two vertices Q and R

2)A set of circles all pass through the points P (1,-3) and Q (5,7). Show that all their centres lie on a straight line and find its equation
2. (Original post by esmeralda123)
Can you help me understand the steps to finding the answers to these type of questions so that I know how to tackle future questions like these:

1) Find k so that point (3, root27) lies on the circle x^2+y^2=k^2. If P, Q and R lie on the circle, and triangle PQR is equilateral, write down the coordinates of the two vertices Q and R

2)A set of circles all pass through the points P (1,-3) and Q (5,7). Show that all their centres lie on a straight line and find its equation
1) You know the centre of the circle, you know the point it has to go through, so the distance between these is the radius - which is your k value. So just find the distance between 2 points.

For the triangle, you can't really determine the two coordinates of Q and R without knowing what P is.

2) For a circle to go through two points, the distance between its centre and each of those two points must be the same.
3. (Original post by RDKGames)
1) You know the centre of the circle, you know the point it has to go through, so the distance between these is the radius - which is your k value. So just find the distance between 2 points.

For the triangle, you can't really determine the two coordinates of Q and R without knowing what P is.

2) For a circle to go through two points, the distance between its centre and each of those two points must be the same.
sorry i didn't make it clear that the point p was (3, root 27)

what's the centre of the circle, where is it given in the question?
4. (Original post by esmeralda123)
sorry i didn't make it clear that the point p was (3, root 27)

what's the centre of the circle, where is it given in the question?
Then in that case you need to find a set of coordinates and such that they satisfy the circle's equation, and

The centre of the circle is given within its equation
5. (Original post by RDKGames)
Then in that case you need to find a set of coordinates and such that they satisfy the circle's equation, and

The centre of the circle is given within its equation
i still don't get how to do it, could you simplify it for me if possible
6. (Original post by esmeralda123)
1) Find k so that point (3, root27) lies on the circle x^2+y^2=k^2.
You need to do some work:
Where is the centre of that circle?
You should now be able to work-out k.

7. (Original post by esmeralda123)
what's the centre of the circle, where is it given in the question?
You have the equation of the circle. The general form of a circle with centre (a,b) and radius r, is:

Now answer the previous questions, so that you know what you're dealing with.

PS: You could just calculate k directly, as you know that satisfies the equation for the circle.
8. (Original post by RogerOxon)
You have the equation of the circle. The general form of a circle with centre (a,b) and radius r, is:

Now answer the previous questions, so that you know what you're dealing with.

PS: You could just calculate k directly, as you know that satisfies the equation for the circle.
would you just need to subtitle the coordinates of P in the equation of the centre of a circle so:

(3-a)^2 + (3root3-b)^2= r^2
9. (Original post by esmeralda123)
i still don't get how to do it, could you simplify it for me if possible
It is much easier and quicker to answer using the rotation matrix since the angle of rotation must be 120 degrees. Otherwise;

You know that the angle between POR, POQ, and ROQ are all 120 as it's an equilateral triangle, and the centre is in the middle of it. You can use the cosine rule to work out the length of the sides for the triangle, let's say this is of value

Then there will be exactly 2 pairs of points which satisfy the equation where represent either of those coordinates.

Also you know the side length, so the distance between P and Q/R is

Then you can rearrange for for one of the variables, plug them back through and find two solutions which will be two different y or x coordinates for your points P and Q - then just plug them back through your derived linear equation in X and Y to find the other axis coordinate for each result that you've just got.

If you follow this slowly and carefully it should make sense. A bit long winded but I don't have time at the moment to look for shorter alternatives - perhaps someone else can help with that.
10. (Original post by esmeralda123)
would you just need to subtitle the coordinates of P in the equation of the centre of a circle so:

(3-a)^2 + (3root3-b)^2= r^2
You have the equation for this circle, so substitute P into that to calculate . Comparison with the general form is used to see where the centre is and what the radius is.
11. (Original post by RogerOxon)
You have the equation for this circle, so substitute P into that to calculate . Comparison with the general form is used to see where the centre is and what the radius is.
3^2 + root 27 ^2 = 36

do i square root 36 to get 6
12. (Original post by esmeralda123)
3^2 + root 27 ^2 = 36

do i square root 36 to get 6
Yes, k=6. (Strictly speaking, it could also be -6, but we'll gloss over that)

Now draw the circle and mark P on it. That should help you visualise it, so that you can see approximately where the other two points are, and calculate their positions - see what RDKGames posted for that.

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