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What is the area of a circle whose equation is (x-3)² + (y + 1) ² =81?
Can someone explain this to me? Im not really looking for solution I want to understand that to do some exercises on my own, please help me with that

Answer :

Cam943
Okay, first, given the equation, we need to find out what the radius of the circle is. Let us state the general equation of a circle: 

[tex] \frac{(x-x_{1})^2}{r^2}+ \frac{(y-y_{1})_^2}{r^2}=1[/tex]

Where [tex](x_{1}, y_{1})[/tex] is the centre of the circle. In this case, we don't need to know the centre. Just the radius. 

Let us start by converting the equation into standard for, which I typed above. Divide both sides by 81. 

[tex] \frac{(x-3)^2}{81}+ \frac{(y+1)^2}{81}=1[/tex]

Great! We now know the radius of the circle. It is [tex] \sqrt{81} [/tex] because it is the bottom fraction. Now we know that the radius is 9. 

So now lets input this into the area of circle formula: 

[tex]A=πr^2[/tex] 

Now we insert our radius. 

[tex]A=9^2π[/tex] 

[tex]=A=81π[/tex] 

You can convert that into a decimal if you wish. 

Hope this helped!

~Cam943, Moderator

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