# Circle problem

1. Jan 5, 2008

### david18

Circle C has equation $x^2+y^2+6x-16=0$

i) find centre and radius (turned out to be centre (-3,0) and radius 5)
ii)verify that point A, (0,4) lies on C
iii)Find the coords of point D, given that AD is a diameter of C

I can do parts i) and ii) and for part iii) would I just use a simultaneous equation with the equation of the line from centre to point A (which worked out as 4/3x+4) and the equation of the circle? Because it seems a bit long winded

2. Jan 5, 2008

### dodo

Point D is at the other end of the circle, with respect to A. How would you express its coordinates?

Last edited: Jan 5, 2008
3. Jan 5, 2008

### david18

I can work it out easily if i sketch the circl;e and it comes out as (-6,-4), but there must be some kind of method to obtain the answer without the sketch.

4. Jan 5, 2008

### dodo

Let me put it this way. Imagine a vector from the center to A. What is the vector from the center to D?

5. Jan 5, 2008

You can subtract two points to obtain a vector:

Make a vector $$\vec{OA}$$ by doing A-O, where O=(0,0), then actually $$\vec{OA}$$ becomes (0,4).

Then let S be the centre of the circle, so S = (-3,0)
And make a vector pointing from A to S $$\vec{AS}$$ = S - A = (-3,-4)

Then just use add vectors like this:
$$\vec{OA}$$ + $$\vec{AD}$$ = $$\vec{OD}$$
but because $$\vec{AD}$$ is diameter it means that:
$$\vec{AD}$$ = 2*$$\vec{AS}$$

So it becomes:
$$\vec{OA}$$ + 2*$$\vec{AS}$$ = $$\vec{OD}$$
From there you have $$\vec{OD}$$ = (-6,-4)
And because $$\vec{OD}$$ = D - O it is the point D, and its coordinates are (-6,-4)

6. Jan 5, 2008

### HallsofIvy

You could use congruent triangles. If A= ($x_1$,$y_1$) is on the circle with center O= ($x_0$,$y_0$), and AB is a diameter, with B= ($x_2$,$y_0$, then The triangle with vertices A, O, and ($x_0$,$y_1$) is congruent to the triangle with vertices B, O, and ($x_0$, $y_1$).

7. Jan 7, 2008

### epenguin

From the centre and radius you have worked out, express the equation of the circle in co-ordinates (x’, y) (you are lucky you have to change only one co-ordinate, the centre lies on the x axis) that have the centre as origin. From your values you know it has got to be (x + 3)^2 + y^2 = 25. Check that this is the same as your original equation. Call it
x’^2 + y^2 = 25

where x’ = (x + 3).

What is the point diametrically opposite to a point A coordinates (x’, y) on a circle centred on origin? Easy -> (-x’, -y)

Now you have that in the new co-ordinates, what is it expressed in the old ones? See last equation.