My Math Forum Standard Equation of Circle HELP!!

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 August 7th, 2016, 06:02 AM #1 Member   Joined: Aug 2016 From: South Korea Posts: 55 Thanks: 0 Standard Equation of Circle HELP!! What is the standard form equation if concentric with x2 + y2 −8x−10y = −16 and 4 times the area ? Explain how you got the answer too please..
 August 7th, 2016, 07:03 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 You do understand that we give HELP, not answers, correct? The general formula for a circle on a Cartesian plane is $(x - h)^2 + (y - k)^2 = r^2$. where (h, k) is the center and r is the radius. Obviously the equation you were given is NOT in the proper form. The general method involved in solving questions like the one you were given is called completing the square. $(u \pm v)^2 = u^2 \pm 2uv + v^2.$ Now suppose you have $u^2 \pm bu.$ That is NOT in the form of a square. You could make it more like the form of the square if you re-expressed it this way: $u^2 \pm bu = u^2 \pm 2 \left( \dfrac{b}{2} \right)u.$ But you need a second squared term to have it in the form of a square. So add zero in the form of the needed square term and its additive inverse. $u^2 \pm 2 \left( \dfrac{b}{2} \right)u = u^2 \pm 2 \left( \dfrac{b}{2} \right)u + 0 = u^2 \pm 2 \left( \dfrac{b}{2} \right)u + \left \{ \left ( \dfrac{b}{2} \right )^2 - \left ( \dfrac{b}{2} \right )^2 \right \}.$ Well I have the extra term I wanted, but I still do not have the form of a square anywhere. So re-arrange the brackets to get the brackets in the right arrangement. $u^2 \pm 2 \left( \dfrac{b}{2} \right)u + \left \{ \left ( \dfrac{b}{2} \right )^2 - \left ( \dfrac{b}{2} \right )^2 \right \} = \left \{u^2 \pm 2 \left( \dfrac{b}{2} \right)u + \left ( \dfrac{b}{2} \right )^2 \right \} - \left ( \dfrac{b}{2} \right )^2.$ Now I have a proper form for a square inside the brackets. $\left \{u^2 \pm 2 \left( \dfrac{b}{2} \right)u + \left ( \dfrac{b}{2} \right )^2 \right \} - \left ( \dfrac{b}{2} \right )^2 = \left ( u \pm \dfrac{b}{2} \right )^2 - \dfrac{b^2}{4}.$ DONE. Thanks from SlayedByMath
 August 7th, 2016, 07:21 AM #3 Member   Joined: Aug 2016 From: South Korea Posts: 55 Thanks: 0 ooh~~ Sorry~~ Im actually collecting some tricky questions on internet about our topic right now on Pre Calculus. And when I actually can't answer a problem that's when I ask for help here Sorry and thanks too
 August 7th, 2016, 07:21 AM #4 Math Team     Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600 Thanks from SlayedByMath
August 7th, 2016, 07:35 AM   #5
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Quote:
 Originally Posted by skeeter Completing the Square: Circle Equations
Oh! Thanks for this!! really need something like this

 August 7th, 2016, 12:34 PM #6 Senior Member   Joined: Jul 2016 From: USA Posts: 108 Thanks: 13 x2 + y2 −8x−10y = −16 Rearrange & add 16 to both sides: x2 −8x + 16 + y2 −10y = 0 Simplify into binomial square. (x - 4)^2 + y2 - 10y = 0 Add 25 to both sides. (x - 4)^2 + y2 - 10y + 25 = 25 Simplify into binomial square! We're Done! (x - 4)^2 + (y - 5)^2 = 25 Not sure if this was the same thing that JeffM1 said but this is the simple algebraic approach. Thanks from SlayedByMath

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