- WELCOME TO A LESSON ON THE CONIC SECTION,
THE CIRCLE.
THE GOALS OF THE VIDEO ARE TO IDENTIFY
THE EQUATION OF A CIRCLE,
WRITE THE STANDARD FORM OF A CIRCLE FROM GENERAL FORM
AND ALSO TO GRAPH A CIRCLE.
LET'S FIRST START OFF BY TALKING ABOUT THE STANDARD FORM
OF AN EQUATION OF A CIRCLE.
A CIRCLE IS THE SET OF POINTS (X, Y)
WHICH ARE A FIXED DISTANCE (R,) THE RADIUS,
AWAY FROM A FIXED POINT (H, K) CALLED THE CENTER.
SO IF THIS IS OUR CIRCLE,
THE CENTER WOULD HAVE THE COORDINATES (H, K).
AND THE DISTANCE FROM THE CENTER
TO ALL OF THE POINTS ON THE CIRCLE
WOULD BE (R) OR THE RADIUS.
NOTICE WHEN THE EQUATION OF THE CIRCLE
IS IN STANDARD FORM
WE CAN EASILY FIND THE (X AND Y) COORDINATE OF THE CENTER
AS WELL AS DETERMINING THE RADIUS.
LET'S GO AHEAD AND TRY A COUPLE EXAMPLES.
SINCE THIS CIRCLE IS IN STANDARD FORM,
WE SHOULD BE ABLE TO IDENTIFY THE CENTER
AND ALSO THE RADIUS.
NOW, NOTICE IT HAS TO BE IN THE FORM OF (X - H) AND (Y - K)
IN ORDER TO FIND THE (X AND Y) COORDINATE OF THIS CENTER.
SO IN THIS CASE THE (X) COORDINATE OF THE CENTER
WOULD BE +3.
AND THE (Y) COORDINATE WOULD BE -2
BECAUSE IF IT HAD TO BE IN THE FORM OF Y MINUS,
IT WOULD BE Y MINUS A -2.
BUT A QUICK EASY WAY TO REMEMBER
THE COORDINATES FOR THE CENTER IS,
IT'S ALWAYS THE OPPOSITE SIGN OF WHAT YOU SEE.
IF YOU SEE -3, IT'S GOING TO BE +3.
IF YOU SEE +2, IT'S GOING TO BE -2.
SO THIS WOULD BE THE CENTER OF THE CIRCLE.
AND THEN (R SQUARED) IS EQUAL TO 16.
IF YOU TAKE THE PRINCIPLE SQUARE ROOT OF BOTH SIDES
THE RADIUS WOULD HAVE LENGTH 4 UNITS.
SO WHAT THAT TELLS US IS IF WE PLOT THE CENTER AT (3, -2)
THE CIRCLE CONSISTS ALL OF THE POINTS
THAT ARE 4 UNITS FROM THIS POINT.
SO WHAT I LIKE TO DO IS GO UP 4 UNITS,
THAT POINT WOULD BE ON THE CIRCLE.
GO DOWN 4 UNITS FROM THE CENTER,
RIGHT 4 UNITS FROM THE CENTER
AND LEFT 4 UNITS FROM THE CENTER.
AND THEN FROM THESE FOUR POINTS TRY TO MAKE A NICE CIRCLE,
SOMETHING LIKE THAT.
SO AS YOU CAN SEE WHEN THE EQUATION IS IN STANDARD FORM,
IT'S PRETTY EASY TO FIND ENOUGH INFORMATION TO MAKE A NICE GRAPH
OF THE CIRCLE.
LET'S GO AHEAD AND TRY ONE MORE.
NOW EVEN THOUGH THIS FORM LOOKS LIKE IT DOESN'T MATCH PERFECTLY
WE COULD REWRITE (X SQUARED) AS THE QUANTITY
X - 0 SQUARE PLUS THE QUANTITY Y - 1 SQUARED = 4.
NOW, THE REASON THAT MIGHT HELP
IS NOW WE CAN PROBABLY DETERMINE MUCH MORE EASILY
THAT THE (X) COORDINATE OF THE CENTER WOULD BE (0)
AND THE (Y) COORDINATE OF THE CENTER WOULD BE (+1).
NEXT, THE RADIUS SQUARED IS EQUAL TO 4.
SO NOW IF WE TAKE THE SQUARE ROOT OF BOTH SIDES
WE HAVE R = 2.
SO WE PLOT THE CENTER (0, 1).
SO THE CIRCLE CONSISTS OF ALL THE POINTS
THAT ARE 2 UNITS FROM THIS POINT IN RED.
SO WE'LL GO UP 2 UNITS, THAT POINT WOULD BE ON THE CIRCLE.
DOWN 2 UNITS, THAT POINT WOULD BE ON THE CIRCLE.
RIGHT 2 UNITS, THAT WOULD BE ON THE CIRCLE.
AND LEFT 2 UNITS,
THAT WOULD ALSO BE ON THE CIRCLE.
SO NOW WE HAVE TO MAKE A CIRCLE
THAT PASSES THROUGH THESE FOUR POINTS,
AND IT WOULD LOOK SOMETHING LIKE THAT.
NOW LET'S TAKE A LOOK AT GENERAL FORM.
IF WE HAVE THE GENERAL FORM OF A CONIC SECTION
IT WILL BE A CIRCLE IF A = C.
MEANING THE COEFFICIENTS OF THE (X SQUARED)
AND THE (Y SQUARED) TERMS MUST EQUAL EACH OTHER,
AND IF THAT'S THE CASE WE HAVE A CIRCLE.
AND WHAT WE CAN DO IS
COMPLETE THE SQUARE ON THE X AND Y PARTS OF THE EQUATION
TO REWRITE THIS IN STANDARD FORM AND THEN GRAPH THE CIRCLE.
SO LET'S GO AHEAD AND GIVE THAT A TRY.
NOTICE THE COEFFICIENT OF THE X SQUARED TERM
AND THE Y SQUARED TERM ARE EQUAL TO EACH OTHER
THEREFORE WE KNOW IT'S A CIRCLE.
BUT SINCE WE HAVE TO COMPLETE THE SQUARE ON THE X PART
AND THE Y PART,
WE WANT THESE LEADING COEFFICIENTS TO BE ONE.
SO WHAT WE'RE GOING TO DO IS DIVIDE EVERY TERM BY 2.
AND, OF COURSE, IF WE DIVIDE 0 BY 2 WE'D STILL HAVE 0.
WHEN WE REWRITE THIS WE'RE GOING TO GROUP THE X TERMS
AND Y TERMS TOGETHER.
SO WE'RE GOING TO HAVE X SQUARED.
THIS WILL BE -6X.
NOW, SINCE WE'RE GOING TO COMPLETE THE SQUARE ON THIS
I'M GOING TO LEAVE ROOM FOR THE CONSTANT TERM
+THE Y SQUARED TERM.
THIS WILL BE +4Y,
AGAIN, LEAVING ROOM FOR THE CONSTANT TERM
WHEN WE COMPLETE THE SQUARE.
NEXT WE'D HAVE -12
OR WE'RE GOING TO ADD 12 TO BOTH SIDES
SO WE HAVE THE CONSTANT TERM ON THE RIGHT SIDE
SO THIS WOULD EQUAL +12.
NEXT, REMEMBER TO COMPLETE THIS SQUARE
WE TAKE HALF OF THIS COEFFICIENT AND THEN SQUARE IT.
HALF OF -6 WILL BE -3 AND -3 SQUARED WOULD BE 9.
SO WE'LL ADD 9 HERE AND THEN,
OF COURSE, WE HAVE TO ADD 9 TO THE RIGHT SIDE AS WELL.
NEXT WE'LL TAKE HALF OF +4,
THAT WOULD BE 2 AND THEN SQUARE IT.
SO WE'RE GOING TO ADD 4 HERE AND ADD 4 TO THE RIGHT.
NOW THESE SHOULD BE PERFECT SQUARE TRINOMIALS.
LET'S TEST IT.
WE WANT THE FACTORS OF 9 THAT ADD TO -6,
THAT WOULD BE -3 AND -3.
SO THIS FACTORS INTO THE QUANTITY X - 3 SQUARE
PLUS WE WANT THE FACTORS OF 4
THAT ADD TO 4 THAT WOULD BE +2 AND +2.
SO THIS WOULD FACTOR INTO THE QUANTITY Y + 2 SQUARED
MUST EQUAL 12 + 9 THAT'D BE 21 + 4 THAT WOULD BE 25.
NOW, THE EQUATION IS IN STANDARD FORM AS WE SEE HERE BELOW.
WE SHOULD BE ABLE TO IDENTIFY THE CENTER AND THE RADIUS.
THE CENTER WOULD BE +3 FOR THE (X) COORDINATE
AND -2 FOR THE (Y) COORDINATE.
AND IF THIS IS R SQUARED THAN R WOULD EQUAL 5.
THIS IS ALL THE INFORMATION WE NEED TO MAKE A NICE GRAPH.
LET'S GO AHEAD AND GRAPH IT ON THE NEXT SCREEN.
THE CENTER IS (3, -2) AND THE RADIUS IS 5.
SO WE'LL GO UP 5 UNITS THAT'D BE +3,
DOWN 5 UNITS WOULD BE -7,
RIGHT 5 UNITS WOULD BE +8
AND LEFT 5 UNITS WOULD BE -2.
AND THEN WE TRY TO MAKE A NICE CIRCLE
PASSING THROUGH THESE FOUR POINTS.
AND THAT'S THE BEST I CAN DO. OKAY.
I HOPE YOU FOUND THIS REVIEW HELPFUL.
THANK YOU FOR WATCHING.