Question

If the area of the circle \( x^2 + y^2 + 8x - 6y + c = 0 \).

Hint:

When finding the area of a circle from its equation, complete the square and find the radius squared.

Answer 1

Step 1: Complete the square.
We begin by completing the square for the given equation of the circle: \[ x^2 + y^2 + 8x - 6y + c = 0 \] First, group the \( x \)-terms and the \( y \)-terms: \[ (x^2 + 8x) + (y^2 - 6y) = -c \] Now, complete the square for both the \( x \)-terms and the \( y \)-terms:
- For \( x^2 + 8x \), add and subtract \( \left( \frac{8}{2} \right)^2 = 16 \).
- For \( y^2 - 6y \), add and subtract \( \left( \frac{-6}{2} \right)^2 = 9 \). The equation becomes: \[ (x^2 + 8x + 16) + (y^2 - 6y + 9) = -c + 16 + 9 \] \[ (x + 4)^2 + (y - 3)^2 = -c + 25 \]
Step 2: Relate to the area of the circle.
The standard equation for a circle is: \[ (x - h)^2 + (y - k)^2 = r^2 \] Here, \( r^2 = -c + 25 \), so the area of the circle is \( \pi r^2 \), which becomes: \[ \text{Area} = \pi(-c + 25) \]