1. CH. 10 DAY 1

2. CH. 10: WE WILL STUDY: CONIC SECTIONS WHEN A PLANE INTERSECTS A RIGHT CIRCULAR CONE, THE RESULT IS A CONIC SECTION. THE FOUR TYPES ARE:

3. CIRCLE Standard Equation if center is the origin Standard Equation if (h,k) is the center x 2 +y 2 = r 2 (x-h) 2 +(y-k) 2 = r 2 r = radius

4. CIRCLE (x-h) 2 +(y-k) 2 = r 2 Find the equation of a circle that has a center =(2,-1) and a radius equal to 3 (x-2) 2 +(y- -1) 2 = 3 2 (x-2) 2 +(y+1) 2 = 9 (x-h) 2 +(y-k) 2 = r 2

5. CIRCLE (x) 2 +(y-3) 2 = 5 2 Find the equation of a circle that has a center = (0,3) and a point on the circumference of the circle is (-4,6) (x) 2 +(y-3) 2 = 25 (3) 2 +(4) 2 = r 2 r = 5 First find the radius! (x-h) 2 +(y-k) 2 = r 2

6. CIRCLE Find the equation of a circle that has (2,-3) and (-10,-8) as the endpoints of a diameter. (x+4) 2 +(y+5.5) 2 =(6.5) 2 Center: (-4,-5.5) d =13 so r = 6.5 First find the center! (5) 2 +(12) 2 = d 2 Now find the radius (x-h) 2 +(y-k) 2 = r 2 (x+4) 2 +(y+5.5) 2 =42.25

7. CIRCLE Given the equation: 4(x-2) 2 +4(y+3) 2 + 20 = 120 A. Write the equation in Standard form B. Find the center of the circle C. Find the radius of the circle (x-h) 2 +(y-k) 2 = r 2 Marker boards 4(x-2) 2 +4(y+3) 2 + 20 = 120 4(x-2) 2 +4(y+3) 2 = 100 A. (x-2) 2 +(y+3) 2 = 25 C. Radius 5 B. Center(2,-3) D. CHALLENGE: If the coefficients on the binomial terms were 4 and 5…how would this impact the equation of the circle? D. It would not be a circle. It would be an ellipse!

8. CHANGE THE EQUATION FROM NON-STANDARD FORM TO STANDARD FORM (x-h) 2 +(y-k) 2 = r 2

9. WRITE THE EQUATION IN STANDARD FORM (x-h) 2 +(y-k) 2 = r 2 To complete the square, send the “c” over the equal. Get the leading coefficient to be 1 by factoring out the 4. Then find the magic number using square of b/2 Can I divide by “4” first thing? In circles – yes – but not in some other conic sections so I wanted to get you ready for those!

10. CIRCLE (x-h) 2 +(y-k) 2 = r 2 Find the equation of a circle, in standard form, that has a center =(-2,-6) and is tangent to the line x=3 (x+2) 2 +(y+6) 2 = 5 2 (x+2) 2 +(y+6) 2 = 25 (x-h) 2 +(y-k) 2 = r 2

11. CIRCLE (x-h) 2 +(y-k) 2 = r 2 Find the equation of a circle, in standard form, that has a center in quadrant 1 and is tangent to the lines x=-3, x=5, and the x-axis (x-1) 2 +(y-4) 2 = 4 2 (x-1) 2 +(y-4) 2 = 16 (x-h) 2 +(y-k) 2 = r 2 Center: (1,4) Domain: Range: Circumference: Area:

12. HW: WS 10.1 DUE NEXT CLASS! HW hints: #3 Divide by 3 first #4 Add 8 first Remember the right side of the equation is r 2 …not r Simplify radicals if you can but do not convert them to decimals!