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Zorluk: OrtaEquations and Graphs of Circles

In the standard (x,y)(x, y) coordinate plane, the equation of a circle is given by 3x2+3y2−12x+24y−15=03x^2 + 3y^2 - 12x + 24y - 15 = 0. What is the radius of this circle?

Cevap: 5

Cevap

The radius of the circle is 5 units.
Dividing the equation 3x2+3y2−12x+24y−15=03x^2 + 3y^2 - 12x + 24y - 15 = 0 by 3 gives x2+y2−4x+8y−5=0x^2 + y^2 - 4x + 8y - 5 = 0. Completing the square for both variables results in (x−2)2+(y+4)2=25(x - 2)^2 + (y + 4)^2 = 25. Because r2=25r^2 = 25, taking the square root gives the radius r=5r = 5.

Adım Adım Çözüm

1
Divide the entire equation by 3.
x2+y2−4x+8y−5=0x^2 + y^2 - 4x + 8y - 5 = 0
The standard form of a circle equation (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2 requires the coefficients of x2x^2 and y2y^2 to be 1.
2
Group terms and move the constant to the right side of the equation.
(x2−4x)+(y2+8y)=5(x^2 - 4x) + (y^2 + 8y) = 5
Isolating quadratic terms makes completing the square straightforward.
3
Complete the square for both variables by adding (−42)2=4(\frac{-4}{2})^2 = 4 and (82)2=16(\frac{8}{2})^2 = 16 to both sides.
(x2−4x+4)+(y2+8y+16)=5+4+16(x^2 - 4x + 4) + (y^2 + 8y + 16) = 5 + 4 + 16
Adding these values turns the expressions into perfect square trinomials while preserving equality.
4
Express in standard form and solve for the radius rr.
(x−2)2+(y+4)2=25  ⟹  r2=25  ⟹  r=5(x - 2)^2 + (y + 4)^2 = 25 \implies r^2 = 25 \implies r = 5
The radius is the principal square root of r2r^2.

Anahtar Kavram

Converting general form of a circle equation to standard form using completing the square.
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Konular

Equations and Graphs of Circles

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