Answer :
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
[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
