Alexis2989127
Answered

Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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.

We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.