Circle in General Form PDF

CIRCLE IN GENERAL FORM The general form of the equation of a circle are: 1. 𝐀𝐱 + 𝐀𝐲 + 𝐃𝐱 + 𝐄𝐲 + 𝐅 = 𝟎 ; where 𝐀 ≠ 𝟎 𝟐 𝟐 2. 𝐱 + 𝐲 + 𝐃𝐱 + 𝐄𝐲 + 𝐅 = 𝟎 𝟐 𝟐 where; 𝐃 = −𝟐𝐡 𝐄 = −𝟐𝐤 𝟐 𝟐 𝟐 𝐅=𝐡 +𝐤 −𝐫 Note: (h, k) is the center and r is t...

CIRCLE IN GENERAL FORM The general form of the equation of a circle are: 1. 𝐀𝐱 + 𝐀𝐲 + 𝐃𝐱 + 𝐄𝐲 + 𝐅 = 𝟎 ; where 𝐀 ≠ 𝟎 𝟐 𝟐 2. 𝐱 + 𝐲 + 𝐃𝐱 + 𝐄𝐲 + 𝐅 = 𝟎 𝟐 𝟐 where; 𝐃 = −𝟐𝐡 𝐄 = −𝟐𝐤 𝟐 𝟐 𝟐 𝐅=𝐡 +𝐤 −𝐫 Note: (h, k) is the center and r is the radius.  Steps: 1. Write in standard form. 2. Expand the (𝐱 − 𝐡) 𝟐 and (𝐲 − 𝐤)𝟐 using foil method. 3. Write the equation in general form. 4. Simplify the expression. 1. Write the equation of the circle in general form with center (2, 1) and radius 3 Solution: The standard form of a circle is 𝟐 𝟐 𝟐 (𝐱 − 𝐡) +(𝐲 − 𝐤) = 𝐫 𝟐 𝟐 (𝐱 − 𝟐) +(𝐲 − 𝟏) = (𝟑)𝟐 𝐱 − 𝟐 (𝐱 − 𝟐) 𝟐 𝟐 𝟐 𝐱 −𝟐𝐱 −𝟐𝐱 +𝟒 (𝐱 − 𝟐) +(𝐲 (𝐲 − 𝟏) = 𝟗 𝐱 𝟐 − 𝟒𝐱 + 𝟒 𝐱 𝟐− 𝟒𝐱 + 𝟒+ 𝐲 𝟐 − 𝟐𝐲 + 𝟏= 𝟗 𝟐 𝟐 𝐱 + 𝐲 + 𝐃𝐱 + 𝐄𝐲 + 𝐅 = 𝟎 𝟐 𝟐 𝐲 − 𝟏 (𝐲 − 𝟏) 𝐱 + 𝐲 − 𝟒𝐱 − 𝟐𝐲 + 𝟒 + 𝟏 = 𝟗 𝟐 𝟐 𝟐 𝐲 − 𝟏𝐲 − 𝟏𝐲 + 𝟏 𝐱 + 𝐲− 𝟒𝐱 − 𝟐𝐲 + 𝟒 + 𝟏 − 𝟗 = 𝟎 𝐲 𝟐 − 𝟐𝐲 + 𝟏 𝟐 𝟐 𝐱 + 𝐲 − 𝟒𝐱 − 𝟐𝐲 − 𝟒 = 𝟎 1. Write the equation of the circle in general form with center (2, 1) and radius 3 Alternative Solution: Since the center is at (2, 1) and r = 3, where h = 2 and k = 1 𝐅= 𝐡 𝟐 + 𝐤 𝟐 − 𝐫 𝟐 𝐃 = −𝟐𝐡 𝐄 = −𝟐𝐤 𝐅= 𝟐 𝟐 + 𝟏 𝟐 − 𝟑 𝟐 𝐃 = −𝟐(𝟐) 𝐄 = −𝟐(𝟏) 𝐃 = −𝟒 𝐄 = −𝟐 𝐅=𝟒+𝟏−𝟗 𝐅 = −𝟒 𝐱 𝟐 + 𝐲 𝟐 + 𝐃𝐱 + 𝐄𝐲 + 𝐅 = 𝟎 𝟐 𝟐 𝐱 + 𝐲 − 𝟒𝐱 − 𝟐𝐲 − 𝟒 = 𝟎 2. Write the equation of the circle in general form with center (-1, -6) and radius 8. Solution: The standard form of a circle is 𝟐 𝟐 𝟐 (𝐱 − 𝐡) +(𝐲 − 𝐤) = 𝐫 (𝐱 − 𝟐 (−𝟏)) +(𝐲 − 𝟐 (−𝟔)) = (𝟖) 𝟐 𝐱 + 𝟏 (𝐱 + 𝟏) 𝟐 𝟐 𝟐 𝐱 +𝟏𝐱 +𝟏𝐱+𝟏 (𝐱 + 𝟏) +(𝐲 (𝐲 + 𝟔) = 𝟔𝟒 𝐱 𝟐 + 𝟐𝐱 + 𝟏 𝐱 𝟐+ 𝟐𝐱 + 𝟏+ 𝐲 𝟐 + 𝟏𝟐𝐲 + 𝟑𝟔 = 𝟔𝟒 𝟐 𝟐 𝐱 + 𝐲 + 𝐃𝐱 + 𝐄𝐲 + 𝐅 = 𝟎 𝟐 𝟐 𝐲 + 𝟔 (𝐲 + 𝟔) 𝐱 + 𝐲 + 𝟐𝐱 + 𝟏𝟐𝐲 + 𝟏 + 𝟑𝟔 = 𝟔𝟒 𝟐 𝟐 𝟐 𝐲 + 𝟔𝐲 + 𝟔𝐲 + 𝟑𝟔 𝒙 + 𝒚 + 𝟐𝐱 + 𝟏𝟐𝐲 + 𝟏 + 𝟑𝟔 − 𝟔𝟒 = 𝟎 𝐲 𝟐 + 𝟏𝟐𝐲 + 𝟑𝟔 𝟐 𝟐 𝐱 + 𝐲 + 𝟐𝐱 + 𝟏𝟐𝐲 − 𝟐𝟕 = 𝟎 2. Write the equation of the circle in general form with center (-1, -6) and radius 8. Alternative Solution: Since the center is at (-1, -6) and r = 8, where h = -1 and k = -6 𝐅= 𝐡 𝟐 + 𝐤 𝟐 − 𝐫 𝟐 𝐃 = −𝟐𝐡 𝐄 = −𝟐𝐤 𝐅 = −𝟏 𝟐 + −𝟔 𝟐 − 𝟖 𝟐 𝐃 = −𝟐(−𝟏) 𝐄 = −𝟐(−𝟔) 𝐃=𝟐 𝐄 = 𝟏𝟐 𝐅 = 𝟏 + 𝟑𝟔 − 𝟔𝟒 𝐅 = −𝟐𝟕 𝐱 𝟐 + 𝐲 𝟐 + 𝐃𝐱 + 𝐄𝐲 + 𝐅 = 𝟎 𝟐 𝟐 𝐱 + 𝐲 + 𝟐𝐱 + 𝟏𝟐𝐲 − 𝟐𝟕 = 𝟎 3. Write the equation of the circle in general form with center (7, -1) and radius 6. Solution: The standard form of a circle is 𝟐 𝟐 𝟐 (𝐱 − 𝐡) +(𝐲 − 𝐤) = 𝐫 𝟐 𝟐 𝟐 (𝐱 − 𝟕) +(𝐲 − (−𝟏)) = (𝟔) 𝟐 𝟐 (𝐱 − 𝟕) +(𝐲 + 𝟏) = 𝟑𝟔 𝟐 𝟐 𝐱 − 𝟏𝟒𝐱 + 𝟒𝟗+ 𝒚 + 𝟐𝐲 + 𝟏 = 𝟑𝟔 𝐱 𝟐 + 𝐲 𝟐 + 𝐃𝐱 + 𝐄𝐲 + 𝐅 = 𝟎 𝐱 𝟐 + 𝐲 𝟐 − 𝟏𝟒𝐱 + 𝟐𝐲 + 𝟒𝟗 + 𝟏 − 𝟑𝟔 = 𝟎 𝐱 𝟐 + 𝐲 𝟐 − 𝟏𝟒𝐱 + 𝟐𝐲 + 𝟏𝟒 = 𝟎  𝟐 𝟐 𝒙 + 𝒚 + 𝑫𝒙 + 𝑬𝒚 + 𝑭 = 𝟎 𝒙−𝒉 𝟐 + 𝒚−𝒌 𝟐 = 𝒓𝟐  Steps: 1. Isolate the constant term. 2. Combine the x and y term. 3. Completing the square for 𝒙 𝐨𝐫 𝒚 , and 𝟐 𝟐 add all the third term to the other side of the equation. 4. Factor the left side using perfect square trinomial. 4. Convert the general equation 𝐱 + 𝐲 + 𝟖𝐱 − 𝟔𝐲 + 𝟐𝟏 = 𝟎 𝟐 𝟐 into standard form equation of a circle. Find the center and radius. Steps: Solution: 1. Isolate the constant term. 𝟐 𝟐 2. Combine the x and y term. 𝐱 + 𝐲 + 𝟖𝐱 − 𝟔𝐲 + 𝟐𝟏 = 𝟎 3. Completing the square for 𝟐 𝟐 𝐱 + 𝐲 + 𝟖𝐱 − 𝟔𝐲 = −𝟐𝟏 𝐱 𝐨𝐫 𝐲 , and add all the third term 𝟐 𝟐 to the other side of the equation. 𝟐 𝟐 (𝐱 +𝟖𝐱) + (𝐲 − 𝟔𝐲) = −𝟐𝟏 4. Factor the left side using perfect square trinomial. 𝟐 (𝐱 +𝟖𝐱 + ____) 𝟐 𝟏𝟔 + (𝐲 − 𝟔𝐲 + ___) 𝟗 = −𝟐𝟏 +𝟏𝟔 +𝟗 𝟐 + 𝐲−𝟑 𝟐 =𝟒 𝐱+𝟒 𝐂 −𝟒 , 𝟑 𝐚𝐧𝐝 𝐫 = 𝟐 5. Convert the general equation 𝟖𝒙 + 𝟖𝒚 − 𝟏𝟔𝒙 + 𝟔𝟒𝒚 − 𝟏𝟎𝟒 = 𝟎 𝟐 𝟐 into standard form equation of a circle. Find the center and radius. Steps: Solution: 1. Isolate the constant term. 𝟖𝒙 𝟐 + 𝟖𝒚𝟐 − 𝟏𝟔𝒙 + 𝟔𝟒𝒚 − 𝟏𝟎𝟒 = 𝟎 2. Combine the x and y term. 𝟖𝒙 𝟐 + 𝟖𝒚𝟐 − 𝟏𝟔𝒙 + 𝟔𝟒𝒚 = 𝟏𝟎𝟒 3. Completing the square for 𝒙 𝐨𝐫 𝒚 , and add all the third term 𝟐 𝟐 𝒙 𝟐 + 𝒚𝟐 − 𝟐𝒙 + 𝟖𝒚 = 𝟏𝟑 to the other side of the equation. 𝟐 𝟐 4. Factor the left side using perfect (𝒙 −𝟐𝒙)+ (𝒚 + 𝟖𝒚) = 𝟏𝟑 square trinomial. 𝟐 𝟐 (𝒙 −𝟐𝒙 + __) 𝟏 + (𝒚 𝟏𝟔 = 𝟏𝟑 +𝟏 +𝟏𝟔 + 𝟖𝒚 + ____) 𝟐+ 𝒚+𝟒 𝟐 = 𝟑𝟎 𝒙−𝟏 𝑪 𝟏 , −𝟒 𝒂𝒏𝒅 𝒓 = 𝟑𝟎 ≈ 𝟓. 𝟒𝟖 6. Convert the general equation + 𝐱 𝟐 𝐲 𝟐 − 𝟐𝐱 − 𝟑𝟓 = 𝟎 into standard form equation of a circle. Find the center and radius. Solution: Steps: 𝟐 𝟐 1. Isolate the constant term. 𝐱 + 𝐲 − 𝟐𝐱 − 𝟑𝟓 = 𝟎 2. Combine the x and y term. 𝐱 𝟐 + 𝐲 𝟐 − 𝟐𝐱 = 𝟑𝟓 3. Completing the square for 𝐱 𝐨𝐫 𝐲 , and 𝟐 𝟐 𝟐 (𝐱 −𝟐𝐱) +𝐲 𝟐 = 𝟑𝟓 add all the third term to the other side of 𝟐 (𝐱 −𝟐𝐱 +𝟏 __) +𝐲 𝟐 = 𝟑𝟓 +𝟏 the equation. 𝟐 + 𝐲 𝟐 4. Factor the left side using perfect square 𝐱−𝟏 = 𝟑𝟔 trinomial. 𝐂 𝟏 , 𝟎 𝐚𝐧𝐝 𝐫 = 𝟔  𝟐 𝟐 𝐀𝐱 + 𝐀𝐲 + 𝐃𝐱 + 𝐄𝐲 + 𝐅 = 𝟎 Or 𝟐 𝟐 𝐱 + 𝐲 + 𝐃𝐱 + 𝐄𝐲 + 𝐅 = 𝟎 STEP 1: Write in standard form. STEP 2: Expand the (𝐱 − 𝐡)𝟐 or (𝐲 − 𝐤)𝟐 using foil method. STEP 3: Write the equation in general form. STEP 4: Simplify the expression. STEP 1: Isolate the constant term. STEP 2: Combine the x and y term. STEP 3: Completing the square for 𝒙 𝐨𝐫 𝒚 , and add all the third 𝟐 𝟐 term to the other side of the equation. STEP 4: Factor the left side using perfect square trinomial.  ACTIVITY 2: MATCH TO SOLVE! A German automobile manufacturer that designs, engineers, produces, markets and distributes luxury vehicles. 1 2 3 4 Direction: Match column A with column B, by determining the standard and general equation of a circle. A B 1. 𝐱 + 𝐲 − 𝟖𝐱 + 𝟐𝐲 + 𝟏𝟑 = 𝟎 𝟐 𝟐 I. 𝟐 𝟐 𝐱+𝟓 + 𝐲+𝟏 =𝟒 2. 𝐱 + 𝐲 − 𝟐𝐱 + 𝟐𝐲 − 𝟐 = 𝟎 𝟐 𝟐 U. 𝟐 𝟐 𝐱−𝟏 + 𝐲+𝟏 =𝟒 3. 𝐱 + 𝐲 + 𝟒𝐱 + 𝟐𝐲 + 𝟏 = 𝟎 𝟐 𝟐 D. 𝟐 𝟐 𝐱+𝟐 + 𝐲+𝟏 =𝟒 4. 𝐱 + 𝐲 + 𝟏𝟎𝐱 + 𝟐𝐲 + 𝟐𝟐 = 𝟎 𝟐 𝟐 A. 𝟐 𝟐 𝐱−𝟒 + 𝐲+𝟏 =𝟒