AI Tutor
If air traffic control receives the radar equation x2+y2βˆ’6x+8y+9=0x^2 + y^2 - 6x + 8y + 9 = 0, where is the storm centered and how far does it reach? In 1637, RenΓ© Descartes connected algebra and geometry, showing that expanded polynomials often conceal clean, symmetric geometric shapes.

From General to Standard Form

A circle's standard form is (xβˆ’h)2+(yβˆ’k)2=r2(x - h)^2 + (y - k)^2 = r^2, which immediately displays its center (h,k)(h, k) and radius rr. The general form, x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0, hides these features by expanding the squares into a single polynomial.
How do we algebraically package those separated xx and yy terms back into neat binomial squares?

Step-by-Step Conversion

To complete the square on x2+bxx^2 + bx, we add (b/2)2(b/2)^2, which rewrites the expression as (x+b/2)2(x + b/2)^2. For x2+y2βˆ’6x+8y+9=0x^2 + y^2 - 6x + 8y + 9 = 0, group variables and move the constant: (x2βˆ’6x)+(y2+8y)=βˆ’9(x^2 - 6x) + (y^2 + 8y) = -9.
Add (βˆ’6/2)2=9(-6/2)^2 = 9 and (8/2)2=16(8/2)^2 = 16 to both sides: (x2βˆ’6x+9)+(y2+8y+16)=βˆ’9+9+16(x^2 - 6x + 9) + (y^2 + 8y + 16) = -9 + 9 + 16. This factors into (xβˆ’3)2+(y+4)2=16(x - 3)^2 + (y + 4)^2 = 16, giving a center at (3,βˆ’4)(3, -4) and a radius of r=16=4r = \sqrt{16} = 4.

Key takeaways

  • The general quadratic equation of a circle in Cartesian space is given by x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0 (or Ax2+Ay2+Dx+Ey+F=0Ax^2 + Ay^2 + Dx + Ey + F = 0, where A=Bβ‰ 0A = B \neq 0 and the cross-term coefficient of xyxy is 00).
  • To convert the general equation into standard (center-radius) form (xβˆ’a)2+(yβˆ’b)2=r2(x - a)^2 + (y - b)^2 = r^2, group xx-terms and yy-terms together, move the constant term to the opposite side: (x2+2gx)+(y2+2fy)=βˆ’c(x^2 + 2gx) + (y^2 + 2fy) = -c.
  • Complete the square independently for both variables by adding the square of half each linear coefficient, (g)2(g)^2 and (f)2(f)^2, to both sides of the equation: (x2+2gx+g2)+(y2+2fy+f2)=g2+f2βˆ’c(x^2 + 2gx + g^2) + (y^2 + 2fy + f^2) = g^2 + f^2 - c.
  • Factor each trinomial into a binomial squared to yield standard form: (x+g)2+(y+f)2=g2+f2βˆ’c(x + g)^2 + (y + f)^2 = g^2 + f^2 - c, identifying the circle's center as (a,b)=(βˆ’g,βˆ’f)(a, b) = (-g, -f) and its radius as r=g2+f2βˆ’cr = \sqrt{g^2 + f^2 - c}.