Question

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

Answer 1

Step-by-step explanation:

The equation of a circle is usually in the form: $(x-h)^2+(y-k)^2=r^2$ where (h, k) is the center and r is the radius. With the information you provided you would have the equation: $(x+3)^2+(y-4)^2=16$. 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:

$(x+3)^2+(y-4)^2=16$

Subtract (x+3)^2 from both sides

$(y-4)^2=-(x+3)^2+16$

Take the square root of both sides

$y-4 = \pm\sqrt{-(x+3)^2+16}$

add 4 to both sides

$y=\pm(\sqrt{-(x+3)^2+16})+4$

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

Answer 2

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: $(x-h)^2 +(y-k)^2=r^2$

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

LEt's plug in the known values

   $(x-(-3))^2+(y-4)^2 = 4^2\\(x+3)^2+(y-4)^2=16$

Answer: (x+3)² + (y-4)² = 16

Hope that helps!

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