At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

Solve the equqtion of a circle (-3,4) and radius 4 units​

Sagot :

Step-by-step explanation:

The equation of a circle is usually in the form: [tex](x-h)^2+(y-k)^2=r^2[/tex] where (h, k) is the center and r is the radius. With the information you provided you would have the equation: [tex](x+3)^2+(y-4)^2=16[/tex]. I'm a bit confused by "solve the equation of a circle" do you mean solve for y? if so you would just do

Original Equation:

[tex](x+3)^2+(y-4)^2=16[/tex]

Subtract (x+3)^2 from both sides

[tex](y-4)^2=-(x+3)^2+16[/tex]

Take the square root of both sides

[tex]y-4 = \pm\sqrt{-(x+3)^2+16}[/tex]

add 4 to both sides

[tex]y=\pm(\sqrt{-(x+3)^2+16})+4[/tex]

and then just change the sign to negative or positive depending on what half of the circle you want

Let's take this problem step-by-step:

I will assume that you want the standard form of the equation of the circle:

      ⇒ standard form: [tex](x-h)^2 +(y-k)^2=r^2[/tex]

  • (h,k) --> coordinate of circle's center --> (-3,4)
  • r --> length of radius --> 4

LEt's plug in the known values

   [tex](x-(-3))^2+(y-4)^2 = 4^2\\(x+3)^2+(y-4)^2=16[/tex]

Answer: (x+3)² + (y-4)² = 16

Hope that helps!

#LearnwithBrainly!