CENTER RADIUS FORM

Question 1

IF THE CENTER IS LOCATED AT THE ORIGIN, WHAT WILL BE THE STANDARF EQUATION OF THE CIRCLE?

Answer 1

C: (0,0)
(X-H)2 + (Y-K)2 = R2
(X-0)2 + (Y-0)2 = R2
X2 + Y2 = R2

Question 2

(X-4)2 + (Y+5)2 = 81

Answer 2

C: ?
(X-4)2 + (Y+5)2 = 81
X-4=0 Y+5=0
X=4 Y=-5 SQUARE ROOT OF 81 R=9

Question 3

IN ORDER TO GET THE GENERAL EQUATION OF A CIRCLE, YOU NEED TO EXPAND THE STANDARD EQUATION.

(X-H)2 + (Y-K)2 = R2

Answer 3

(X-H)2 + (Y-K)2 = R2
(X2 - 2HX + H2) + (Y2 - 2YK + K2) = R2
X2 - 2HX + H2 + Y2 - 2YK + K2 = R2
X2 - 2HX + H2 + Y2 - 2YK + K2 - R2 = 0

Question 4

HOW TO KNOW IF IT IS AN EQUATION OF A CIRCLE ?

Answer 4

> BOTH X AND Y HAVE THE HIGHEST DEGREE OF 2.
BOXT X2 AND Y2 HAVE THE SMAE NUMRICAL COEFFICIENT.
THE OPERATION BETWEEN X2 AND Y2 IS ADDITION.

Question 5

H DENOTES THE …

Answer 5

X-COORDINATE

Question 6

K DENOTES THE …

Answer 6

Y-COORDINATE

Question 7

IN ORDER TO SIMPLIFY YOUR EQUATION PARTICULARLY YOUR SOLUTION.

Answer 7

X2 + Y2 + DX + EY + F = 0