1. The problem is to understand the key concepts and formulas related to circles in applied mathematics. 2. The standard equation of a circle with center at $(h,k)$ and radius $r$ is given by: $$ (x - h)^2 + (y - k)^2 = r^2 $$ This formula is fundamental for solving problems involving circles. 3. Important rules: - The center $(h,k)$ is the point from which all points on the circle are equidistant. - The radius $r$ is the distance from the center to any point on the circle. 4. Example: Find the equation of a circle with center at $(3,-2)$ and radius $5$. 5. Substitute $h=3$, $k=-2$, and $r=5$ into the formula: $$ (x - 3)^2 + (y + 2)^2 = 5^2 $$ 6. Simplify the radius squared: $$ (x - 3)^2 + (y + 2)^2 = 25 $$ 7. This is the equation of the circle. 8. Another important concept is the general form of a circle's equation: $$ x^2 + y^2 + Dx + Ey + F = 0 $$ where $D$, $E$, and $F$ are constants. 9. To find the center and radius from the general form, complete the square for $x$ and $y$ terms. 10. Example: Convert $$ x^2 + y^2 - 6x + 8y + 9 = 0 $$ to standard form. 11. Group $x$ and $y$ terms: $$ (x^2 - 6x) + (y^2 + 8y) = -9 $$ 12. Complete the square: - For $x$: take half of $-6$ is $-3$, square it to get $9$. - For $y$: take half of $8$ is $4$, square it to get $16$. 13. Add $9$ and $16$ to both sides: $$ (x^2 - 6x + 9) + (y^2 + 8y + 16) = -9 + 9 + 16 $$ 14. Simplify: $$ (x - 3)^2 + (y + 4)^2 = 16 $$ 15. The center is $(3, -4)$ and the radius is $\sqrt{16} = 4$. This revision covers the fundamental formulas and methods for working with circles in applied mathematics.