Question

A circle is represented by the equation x2 + 8x + y2 – 12y = –29. Describe the algebraic technique(s) needed to rewrite the equation in standard form. Identify the center and radius of the circle. Show your work

Answer 1

The equation of the circle can be written in standard form and in a general form

The center of the circle is (-4,6) and the radius is $\sqrt{23$

How to identify the radius and the center of the circle

The equation of the circle is given as:

$x^2 + 8x + y^2 - 12y = -29$

Rewrite as:

$x^2 + 8x + k - k + y^2 - 12y + k - k= -29$

Express k as the square of the half of the coefficients of x and y.

So, we have:

$x^2 + 8x + (4)^2 - (4)^2 + y^2 - 12y + (6)^2 - (6)^2= -29$

Evaluate the exponents

$x^2 + 8x + 16 - 16 + y^2 - 12y + 36 - 36 = -29$

Collect like terms

$x^2 + 8x + 16 + y^2 - 12y + 36 = -29 + 16 + 36$

Express as perfect squares

$(x + 4)^2 + (y - 6)^2 = -29 + 16 + 36$

$(x + 4)^2 + (y - 6)^2 = 23$

The standard equation of a circle is:

$(x - a)^2 + (y - b)^2 + r^2$

Where:

Center = (a,b)

Radius = r

So, we have:

Center = (-4,6)

Radius = $\sqrt{23$

Hence, the center of the circle is (-4,6) and the radius is $\sqrt{23$

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