- WE ARE ASKED TO WRITE THE CENTERED EQUATION OF A CIRCLE
WITH A DIAMETER WITH END POINTS (-2,1) AND (4,3).
THE CENTERED EQUATION OF A CIRCLE IS GIVEN HERE
WHERE (H,K) ARE THE COORDINATES OF THE CENTER,
AND R IS THE LENGTH OF THE RADIUS.
SO KNOWING THE TWO ENDPOINTS OF THE DIAMETER
WHICH IS A SEGMENT THAT PASSES THROUGH THE CENTER OF THE CIRCLE
AND HAS TWO END POINTS ON THE CIRCLE,
WE WANT TO FIND THE CENTER OF THE CIRCLE
AND THE LENGTH OF THE RADIUS.
TO UNDERSTAND HOW TO APPROACH THIS,
LET'S TAKE A LOOK AT THE CIRCLE THAT HAS THESE TWO ENDPOINTS
FOR A DIAMETER.
HERE'S THE DIAMETER WE WERE REFERRING TO.
NOTICE HOW IT HAS ONE ENDPOINT AT (-2,1),
AND ANOTHER ENDPOINT AT (4,3).
AGAIN, OUR GOAL HERE IS TO FIND THE COORDINATES OF THE CENTER
WHICH WOULD BE (H,K), AND THEN THE LENGTH OF THE RADIUS
WHICH WOULD BE THE LENGTH OF THIS SEGMENT HERE.
NOTICE HOW THE CENTER OF THE CIRCLE WOULD
BE THE MIDPOINT OF THE DIAMETER.
AND THEN ONCE WE FIND THE MIDPOINT
OR THE CENTER OF THE CIRCLE,
WE CAN THEN FIND THE LENGTH OF THE RADIUS
BY DETERMINING THE DISTANCE FROM THE CENTER
TO ONE OF THE TWO POINTS ON THE CIRCLE.
SO LET'S GO AHEAD AND DO THAT.
AND, AGAIN, THE CENTER WOULD HAVE COORDINATES (H,K)
WHICH WOULD BE THE MIDPOINT OF THE DIAMETER.
THE MIDPOINT FORMULA IS GIVEN HERE FOR REVIEW.
WE'RE BASICALLY GOING TO FIND THE AVERAGE OF THE X COORDINATES
AND THE AVERAGE OF THE Y COORDINATES.
LET'S CALL THESE COORDINATES THE 1s, X SUB 1 AND Y SUB 1.
AND THESE COORDINATES THE 2, X SUB 2, Y SUB 2.
SO WE WOULD HAVE -2 + 4 DIVIDED BY 2 AND 1 + 3 DIVIDED BY 2.
WELL, -2 + 4 = 2.
2 DIVIDED BY 2 IS 1.
AND 1 + 3 = 4.
4 DIVIDED BY 2 IS 2.
SO NOW WE KNOW THAT H = 1 AND K = 2.
AND NOW TO FIND THE LENGTH OF THE RADIUS
WE'LL FIND THE DISTANCE FROM THE CENTER
TO ONE OF THE ENDPOINTS OF THE DIAMETER.
SO LET'S GO AHEAD AND USE THE CENTER
AND THIS FIRST ENDPOINT (-2,1).
SO THE LENGTH OF THE RADIUS, OR R, IS GOING TO BE EQUAL
TO THE DISTANCE BETWEEN THESE TWO POINTS
AND THE DISTANCE FORMULA IS GIVEN HERE.
SO LET'S GO AHEAD AND CALL THESE THE 1s,
X SUB 1 AND Y SUB 1,
AND THESE THE 2s, X SUB 2, Y SUB 2.
SO THE LENGTH OF THE RADIUS OR THE DISTANCE
IS GOING TO BE EQUAL TO X SUB 2 - X SUB 1,
OR -2 - 1 SQUARED + Y SUB 2 - Y SUB 1 OR 1 - 2 SQUARED.
SO WE'LL HAVE THE SQUARE ROOT OF -3 SQUARED.
THAT'S 9 PLUS (1 - 2) = -1.
-1 SQUARED IS 1.
SO THE RADIUS IS EQUAL TO THE SQUARE ROOT 10.
BUT FOR THE STANDARD EQUATION OF THE CIRCLE
WE NEED R SQUARED, NOT R.
SO IF WE KNOW R EQUALS SQUARE ROOT 10,
WE CAN FIND R SQUARED
BY SQUARING BOTH SIDES OF THE EQUATION.
AND THE SQUARE ROOT OF 10 SQUARED WOULD JUST BE 10.
NOW THAT WE HAVE (H,K) AND R SQUARED
WE CAN WRITE THE EQUATION IN STANDARD FORM.
WE WOULD HAVE THE QUANTITY X - H WHERE H IS 1.
SO WE HAVE X - 1 SQUARED + (Y - K) WHERE K IS 2.
SO WE HAVE Y - 2 SQUARED = R SQUARED WHERE WE KNOW R = 10.
THIS WOULD BE THE STANDARD EQUATION OF THE CIRCLE
WITH THE DIAMETER WITH THE GIVEN ENDPOINTS.
I HOPE YOU FOUND THIS EXPLANATION HELPFUL.