Ex 1: Find Standard Equation of a Circle Given the Endpoints of a Diameter

Ex 1: Find Standard Equation of a Circle Given the Endpoints of a Diameter


– 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.  

21 thoughts on “Ex 1: Find Standard Equation of a Circle Given the Endpoints of a Diameter

  • October 14, 2013 at 9:30 pm
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    Thank you:)

    Reply
  • January 17, 2014 at 1:25 am
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    Thanks a million

    Reply
  • January 31, 2014 at 8:11 pm
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    does it matter which endpoint you use? if I used (4,3) instead of (-2, 1) would it make a difference?

    Reply
  • February 6, 2014 at 4:59 am
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    Thanks! That was really helpful!

    Reply
  • March 5, 2014 at 4:19 am
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    Thankyouthankyouthankyou

    Reply
  • May 26, 2014 at 12:18 am
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    your awesome  thanks so much

    Reply
  • September 24, 2014 at 11:37 pm
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    Literally your amazing. explained this so well 🙂

    Reply
  • October 14, 2014 at 4:53 am
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    Thank You!!! It was very helpful

    Reply
  • May 26, 2015 at 6:51 pm
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    This helped me understand the other parts. I thought you stop at the first part but i see now that there are more steps towards this. Thanks a bunch

    Reply
  • October 10, 2015 at 6:55 am
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    Thank you for posting this video 🙂

    Reply
  • February 4, 2016 at 2:44 am
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    great thankssssss

    Reply
  • May 11, 2016 at 1:35 am
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    Thank you!

    Reply
  • June 8, 2016 at 6:53 pm
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    thank you so much for this video. it really made things clearer and easier to understand☺☺☺

    Reply
  • August 31, 2016 at 10:06 pm
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    This video really helped me! Very clear and easy to understand. Thank you very much for sharing.

    Reply
  • January 29, 2017 at 9:32 pm
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    Thanks a million times you saved me on my report today

    Reply
  • March 20, 2017 at 11:33 pm
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    XD you are my man..thx a lot…

    Reply
  • August 24, 2017 at 4:54 pm
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    Awesome! this also works for spheres if you add a third dimension of z. Thanks helped out in my Calculus 3 class..

    Reply
  • January 28, 2018 at 7:00 am
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    Thank you thank you thank you thank you thank you. Keep making a lot of this… also make a programming tutorial with clear explaination like this thank you

    Reply
  • June 21, 2018 at 10:08 am
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    tysm for this! x.

    Reply
  • May 16, 2019 at 2:06 am
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    is this algebra 1 or 2

    Reply
  • September 25, 2019 at 7:03 pm
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    thank you!!! your video broke this down enough for the problem to make sense! take my thumbs up and subscription!

    Reply

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