The equation x^2 + y^2 - 12x + 8y + 3 = 0 defines a circle in the standard (x, y) coordina… — ACT

The equation $x^2 + y^2 - 12x + 8y + 3 = 0$ defines a circle in the standard $(x, y)$ coordinate plane. What is the radius of this circle?

Cevap: 7

The radius of the circle is 7.

Completing the square on the given equation yields

(x - 6)^2 + (y + 4)^2 = 49

. By comparing this to the standard circle equation

(x - h)^2 + (y - k)^2 = r^2

, we find

r^2 = 49

. Taking the square root of

49

gives a radius of

7

Group the

x

and

y

terms and move the constant to the right-hand side of the equation.

x^2 - 12x + y^2 + 8y = -3

This groups terms containing the same variables together to prepare for completing the square.

Add the square of half of the linear coefficients to both sides to complete the square for both

x

and

y

(x^2 - 12x + 36) + (y^2 + 8y + 16) = -3 + 36 + 16

Adding

(\frac{-12}{2})^2 = 36

and

(\frac{8}{2})^2 = 16

to both sides forms perfect square trinomials on the left side while maintaining equality.

Factor the trinomials into squared binomials and combine the constants on the right side.

(x - 6)^2 + (y + 4)^2 = 49

This rewrites the equation in the standard circle equation form.

Extract the radius from the standard form equation

(x - h)^2 + (y - k)^2 = r^2

r = 7

Since the constant on the right side corresponds to

r^2

, taking the square root of

49

gives the radius of the circle.

Converting a circle's equation from general form to standard form by completing the square to determine its properties.

Soru