What is the area of the circle defined by $x^2-6x +y^2-14y +33=0$ that lies beneath the line $y=7$?
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:
Let's think step by step.

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\frac{25\pi}{2}
What is the area of the circle defined by $x^2-6x +y^2-14y +33=0$ that lies beneath the line $y=7$?
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