Determine the centre and radius of a circle whose equation is 1/2x(4x-8)+ 1/4y(8y−12)=24

The equation is $1/2x(4x-8) + 1/4y(8y-12) = 24$.
Expand the terms:
$(1/2x)(4x) - (1/2x)(8) + (1/4y)(8y) - (1/4y)(12) = 24$
$2x^2 - 4x + 2y^2 - 3y = 24$

Divide the entire equation by 2:
$x^2 - 2x + y^2 - (3/2)y = 12$

Group the x-terms and y-terms:
$(x^2 - 2x) + (y^2 - (3/2)y) = 12$

To complete the square for $x^2 - 2x$, take half of the coefficient of x (-2), which is -1, and square it $(-1)^2 = 1$.
To complete the square for $y^2 - (3/2)y$, take half of the coefficient of y (-3/2), which is -3/4, and square it $(-3/4)^2 = 9/16$.

Add these values to both sides of the equation:
$(x^2 - 2x + 1) + (y^2 - (3/2)y + 9/16) = 12 + 1 + 9/16$

$(x - 1)^2 + (y - 3/4)^2 = 13 + 9/16$

Calculate the right side:
$13 + 9/16 = (13 \cdot 16)/16 + 9/16 = 208/16 + 9/16 = 217/16$

So the equation in standard form is:
$(x - 1)^2 + (y - 3/4)^2 = 217/16$

Comparing this to $(x-h)^2 + (y-k)^2 = r^2$:
The center (h, k) is $(1, 3/4)$.
The radius squared $r^2$ is $217/16$.
The radius $r$ is sqrt(217/16) = sqrt(217) / sqrt(16) = sqrt(217) / 4.