1. ## circle

2. i just need to know how to start it, the second problem th answer for the slop cannot contain T and S.

3. Originally Posted by midnightwynter
i just need to know how to start it, the second problem th answer for the slop cannot contain T and S.
I've modified the first sketch a little bit.

If the center of the circle has the coordinates $\displaystyle C(x_C, y_C)$ then the point P has the coordinates $\displaystyle P(x_C+u , y_C + v)$

Use the right triangle with the legs u and v and the hypotenuse r . Then
$\displaystyle \color{red}v = r \cdot \sin(q)$ and
$\displaystyle \color{blue}u = r \cdot \cos(q)$

4. I've modified the 2nd sketch a little bit.

1. Determine the coordinates of S and T. Use the attached sketch.

2. You are supposed to know the formula to calculate the slope of a line which runs through 2 points.

If the points are $\displaystyle P(x_P, y_P)$ and $\displaystyle Q(x_Q, y_Q)$ then the slope, determined by these points, is:

$\displaystyle m_{PQ}=\dfrac{y_Q - y_P}{x_Q - x_P}$

5. Originally Posted by earboth
I've modified the first sketch a little bit.

If the center of the circle has the coordinates $\displaystyle C(x_C, y_C)$ then the point P has the coordinates $\displaystyle P(x_C+u , y_C + v)$

Use the right triangle with the legs u and v and the hypotenuse r . Then
$\displaystyle u = r \cdot \sin(q)$ and
$\displaystyle v = r \cdot \cos(q)$
Excuse me but I believe you have that backwards: u= r cos(q) and v= r sin(q).

6. Originally Posted by HallsofIvy
Excuse me but I believe you have that backwards: u= r cos(q) and v= r sin(q).
Of course you are right. Thanks for spotting my mistake.

I've "repaired" this accident.