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Can someone please please tell me what the general form for (x-6)^2+(y-3)^2=16 is. I would really appreciate the help!

Sagot :

Answer:

x² + y² - 12x - 6y + 29 = 0

Step-by-step explanation:

Simplifying the equation using (a + b)² = a² + 2ab + b²:

(x - 6)² + (y - 3)² = 16

⇒ [x² - 2(x)(6) + 6²] + [y² - 2(y)(3) + 3²] = 16

⇒ [x² - 12x + 36] + [y² - 6y + 9] = 16

⇒ x² - 12x + 36 + y² - 6y + 9 = 16

General form of a circle = x² + y² + Cx + Dy + E = 0

Before we reorganize the equation in general form, we need to have the R.H.S as 0. For that, we need to subtract 16 both sides.

Subtract 16 both sides:

⇒ x² - 12x + 36 + y² - 6y + 9 = 16

⇒ x² - 12x + 36 + y² - 6y + 9 - 16 = 16 - 16

⇒ x² - 12x + 20 + y² - 6y + 9 = 0

Reorganizing the equation in general form:

x² - 12x + 20 + y² - 6y + 9 = 0

⇒ x² - 12x + 20 + y² - 6y + 9 = 0

⇒ x² + y² - 12x + 20 - 6y + 9 = 0

⇒ x² + y² - 12x + 20 - 6y + 9 = 0

⇒ x² + y² - 12x - 6y + 20 + 9 = 0

⇒ x² + y² - 12x - 6y + 29 = 0

Thus, the equation in general form is x² + y² - 12x - 6y + 29 = 0.

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