Alicia's Reflection #17

Alrighty well since our exam is on all of chapters 1-9, im going to reflect on some old stuff we learned in the beginning of the year.

CIRCLES:

The standard equation of a circle is (x-h)^2+(y-k)^2 .....the center is (h,k)

If the equation is not in standard form, you must complete the square to put it in standard form.

If you are given a center and a point, you can use the distance formula to find the radius.

To find the intersection of a line and a circle:

1. solve the linear eqn for y.
2. substitute in the circle eqn.
3. solve for x.
4. plug the x value in to get the y value.

***Reminder. If your x value is imaginary, then there is no point of intersection.

EX: find the center and radius.

(x-3)^2+(y+7)^2=19

c:(h,k)

center: (3,-7)

radius: square root of 19

EX: find the eqn of the circle with the center (1,4) through (3,7)

--in the problem you are given a center and a point so you would plug into the distance formula.

square root of (3-1)^2+(7-4)^2= square root of 4+9=square root of 13.

**13 has no root.

Your answer should be (x-1)^2+(y-4)^2=13

I also can use some help with finding integral coefficients. I had no clue how to do it on the study guides. Thankss!!