Question

What is the area of the circle defined by x²−6x+y²−14y+33=0 that lies beneath the line y=7?

A) 17π square units
B) 33π square units
C) 50π square units
D) 66π square units

Answer 1

To find the area of the circle defined by the equation x²−6x+y²−14y+33=0 that lies beneath the line y=7, rewrite the equation in standard form, which gives (x-3)² + (y-7)² = 1. This equation represents a circle with center (3, 7) and radius 1. Since the line y=7 intersects the circle at its center, the area of the circle beneath the line y=7 is equal to half the area of the circle, which is 0.5π square units.

Step-by-step explanation:

To find the area of the circle defined by the equation x²−6x+y²−14y+33=0 that lies beneath the line y=7, we need to first rewrite the equation in standard form. Completing the square for both x and y, the equation can be rewritten as (x-3)² + (y-7)² = 1, which is the equation for a circle with center (3, 7) and radius 1. Since the line y=7 intersects the circle at its center, the area of the circle beneath the line y=7 is equal to half the area of the circle, which is 0.5π square units.