IDNLearn.com: Your one-stop destination for finding reliable answers. Get accurate and comprehensive answers from our network of experienced professionals.
Sagot :
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
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Thank you for trusting IDNLearn.com. We’re dedicated to providing accurate answers, so visit us again for more solutions.