Tricky Circles (Mistake Fixed)

The distance between two points is given $d=\sqrt{(x_2-x_1)^2(y_2-y-1)^2}$ . So the length of the radius of the circle is $r=\sqrt{(4-2)^2+(7-9)^2}=\sqrt{4+4}=\sqrt{4(2)}=2\sqrt{2}$ . Finally, the value of $\dfrac{\mbox{Area}}{2 \pi}=\dfrac{\pi (2\sqrt{2})^2}{2 \pi}=$ $\color{#3D99F6}\boxed{\large 4}$

By the distance formula , we get

$r=\sqrt{(4-2)^2+(7-9)^2}=2\sqrt{2}$

The area of the circle is $\pi (2\sqrt{2})^2=\pi (4)(2)=8 \pi$ . Dividing the area by $2 \pi$ , we get an answer of $\color{#D61F06}\boxed{\large 4}$ .

It should be correct this time....

Solution: Find the length of the radius: $\sqrt{(4-2)^2+(7-9)^2}=\sqrt 8$ . Then find the area of the circle: $A=\pi r^2=\pi\times\sqrt 8^2=8\pi$ . Divide by $2\pi$ and get $4$