- WELCOME TO AN EXAMPLE
WHERE WE'RE GIVEN THE GENERAL EQUATION OF A CIRCLE
AND WE'RE ASKED TO WRITE THE EQUATION IN STANDARD FORM
AND THEN GIVE THE CENTER AND RADIUS OF THE CIRCLE.
WE KNOW THIS EQUATION IS A CIRCLE
BECAUSE THE (X) SQUARED AND (Y) SQUARED TERMS
HAVE THE SAME COEFFICIENT.
IN THIS CASE THEY'RE BOTH (1).
SO WE WANT TO WRITE THE EQUATION HERE,
IN GENERAL FORM,
INTO STANDARD FORM WHICH IS GIVEN HERE BELOW.
THE QUANTITY (X) - (H) SQUARED
+ THE QUANTITY (Y) - (K) SQUARED = (R) SQUARED.
IT'S THE STANDARD FORM OF A CIRCLE
WHERE IN THIS FORM THE CENTER HAS COORDINATES (H, K)
AND THE RADIUS IS EQUAL TO (R).
SO THE QUANTITY (X) - (H) SQUARED
MEANS WE HAVE A PERFECT SQUARE TRINOMIAL IN TERMS OF (X).
AND THE QUANTITY (Y) - (K) SQUARED
MEANS WE HAVE A PERFECT SQUARE TRINOMIAL
IN TERMS OF (Y) AS WELL.
AND WE'LL HAVE TO FORM THESE PERFECT SQUARE TRINOMIALS
USING THE GENERAL FORM OF OUR EQUATION.
SO WE'LL HAVE (X) SQUARED - 6X PLUS SOME CONSTANT
THAT WOULD MAKE THIS A PERFECT SQUARE TRINOMIAL
PLUS (Y) SQUARED PLUS 8Y PLUS A CONSTANT
THAT WILL MAKE THIS A PERFECT SQUARE TRINOMIAL.
LET'S GO AHEAD AND MOVE THE CONSTANT FIVE
TO THE RIGHT SIDE SO WE'LL HAVE = -5.
BUT BECAUSE WE'RE GOING TO BE ADDING TO CONSTANTS ON THE LEFT
LET'S GO AHEAD AND PUT + BLANK + BLANK
TO REMIND US TO ADD THE CONSTANT TO BOTH SIDES OF THE EQUATION.
AND, NOW, TO FIND THE CONSTANT
TO MAKE THIS A PERFECT SQUARE TRINOMIAL
WE'RE GOING TO TAKE HALF OF THE COEFFICIENT OF X AND SQUARE IT.
SO IF WE TAKE -6 AND DIVIDE BY 2 AND SQUARE IT THIS VALUE
WILL MAKE THIS A PERFECT SQUARE TRINOMIAL.
WELL, -6 DIVIDED BY 2 IS -3.
-3 SQUARED IS EQUAL TO +9.
SO WE'RE GOING TO ADD 9 HERE
AND ALSO ADD NIGHT TO THE RIGHT SIDE OF THE EQUATION.
AND, NOW, TO MAKE THIS SECOND TRINOMIAL A PERFECT SQUARE
WE'LL TAKE HALF OF 8 AND SQUARE
IT WHICH WOULD BE 4 SQUARED WHICH IS EQUAL TO 16.
SO WE'LL ADD 16 HERE AND WE'LL ADD 16
TO THE RIGHT AS WELL.
NOW WE'LL FACTOR THE X PART AND THE Y PART SEPARATELY
AND THEY SHOULD BE PERFECT SQUARE TRINOMIALS.
SO WE'LL SHOW ALL THE WORK IN THIS EXAMPLE.
THIS WILL FACTOR INTO TWO BINOMIALS WHERE WE HAVE X AND X,
THE FACTORS OF +9 THAT ADD TO - 6 WOULD BE -3 AND -3.
SO WE HAVE X -3 AND X -3.
NOTICE HOW WE CAN REWRITE THIS AS THE QUANTITY X -3 SQUARED
WHICH WE'LL DO IN THE NEXT STEP.
PLUS THE TRINOMIAL IN TERMS OF Y WILL FACTOR INTO TWO BINOMIALS.
SO WE'LL HAVE Y AND Y.
AND NOW THE FACTORS OF +16 THAT ADD TO +8 ARE 4 AND 4.
SO WE'LL HAVE Y + 4 AND Y + 4.
NOTICE ON THE RIGHT WE HAVE -5 + 9 THAT'S 4 + 16 IS 20.
AND NOW WE CAN WRITE THIS IN STANDARD FORM.
WE HAVE THE QUANTITY
X - 3 SQUARED + THE QUANTITY Y + 4 SQUARED = 20.
NOW THAT WE HAVE THIS IN STANDARD FORM
WE CAN FIND THE CENTER AND THE RADIUS.
BECAUSE WE HAVE THE QUANTITY
X - 3 SQUARED THE (X) COORDINATE IS (+3)
AND BECAUSE THE QUANTITY
Y + 4 SQUARED THE (Y) COORDINATE IS (-4).
SO NOTICE HOW THE SIGNS OF THESE COORDINATES
ARE THE OPPOSITE OF WHAT WE SEE HERE
BECAUSE OF THE STANDARD FORM OF A CIRCLE.
FOR Y + 4 WE COULD WRITE THIS AS Y - -4
BUT IT'S EASIER JUST TO REMEMBER
THAT THE SIGNS ARE THESE COORDINATES
ARE THE OPPOSITE OF WHAT WE SEE HERE.
IF WE HAVE X - 3 THE (X) COORDINATE IS +3.
IF WE HAVE Y + 4 THE (Y) COORDINATE IS -4.
AND THEN TO FIND THE RADIUS THE RIGHT SIDE OF THIS EQUATION
IN STANDARD FORM IS EQUAL TO R SQUARED
WHICH MEANS R SQUARED IS EQUAL TO 20.
THIS REPRESENTS THE LENGTH.
SO WE CAN TAKE THE PRINCIPAL SQUARE ROOT
OR POSITIVE SQUARE ROOT OF BOTH SIDES TO FIND (R).
SO WE'D HAVE R = SQUARE ROOT 20 WHICH DOES SIMPLIFY.
20 IS EQUAL TO 4 x 5.
SO WE CAN WRITE R IS EQUAL TO 2 SQUARE ROOT 5
OR IF I WANT A DECIMAL APPROXIMATION
THIS WOULD BE APPROXIMATELY 4.47.
SO WE'LL GO AHEAD AND GIVE THE EXACT VALUE.
THAT WOULD BE 2 SQUARE ROOT OF 5 UNITS
BUT IN ORDER TO GRAPH THIS CIRCLE BY HAND
THE DECIMAL APPROXIMATION WOULD PROBABLY BE MORE USEFUL.
SO, AGAIN, THIS QUESTION ASKS US
FOR THE STANDARD FORM OF THE CIRCLE
AND THE COORDINATES OF THE CENTER
AND THE LENGTH OF THE RADIUS.
SO THIS INFORMATION IS RIGHT HERE.
AND WE'LL GO AHEAD AND FINISH BY GRAPHING THIS.
HERE IS OUR CIRCLE.
NOTICE THE CENTER HAS COORDINATES (3, -4)
IF WE SKETCH THE SEGMENT FROM THE CENTER
TO A POINT IN THE CIRCLE, LET'S SAY THIS SEGMENT HERE,
THIS WOULD BE THE RADIUS
WHICH WE FOUND TO BE 2 SQUARE ROOT 5 OR APPROXIMATELY 4.47.
OKAY, I HOPE YOU FOUND THIS EXPLANATION HELPFUL.