Question

A circle in the xyxyx, y-plane has the equation 36x^2+36y^2-12x-27y-8=036x 2 +36y 2 −12x−27y−8=036, x, start superscript, 2, end superscript, plus, 36, y, start superscript, 2, end superscript, minus, 12, x, minus, 27, y, minus, 8, equals, 0. how long is the radius of the circle?

Answer 1

we have

$36x^2+36y^2-12x-27y-8=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(36x^2-12x)+(36y^2-27y)=8$

Factor the leading coefficient of each expression

$36*(x^2-(1/3)x)+36*(y^2-(3/4)y)=8$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$36*(x^2-(1/3)x+(1/36))+36*(y^2-(3/4)y+(9/64))=8+1+5.0625$

$36*(x^2-(1/3)x+(1/36))+36*(y^2-(3/4)y+(9/64))=14.0625$

Rewrite as perfect squares

$36*(x-(1/6))^2+36*(y-(3/8))^2=14.0625$

Divide both sides by 36

$(x-(1/6))^2+(y-(3/8))^2=0.390625$

$(x-(1/6))^2+(y-(3/8))^2=0.625^2$

the radius of the circle is equal to

r=0.625 units

the answer is

the radius is equal to r=0.625 units