Question

What is the area of the circle defined by x^2-6x +y^2-14y +33=0 that lies beneath the line y=7?

Answer 1

Answer:

$25\pi/2$

Step-by-step explanation:

We complete the square for the circle standard form. $x^2-6x +9+y^2-14y+49 +33=0+9+49$

Simplifying this

$(x-3)^{2} + (x-7)^{2} = 5^2$

We see that the vertex of the circle is (3, 7), since we divide the circle at  y = 7.  This divides the circle into a semi-circle (half of the real circle). Since the radius is 5 we calculate the area of the big circle which is $25\pi$. Half of $25\pi$ is $25\pi/2$.

Thus getting your awnser