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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]
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!
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