If we graph the circle we can see it has the origin (0,0) as its centre. If we draw a line from the origin to the given point, this is a radius of the circle length 2sqrt(6). The tangent is at right angles to the radius. What is the slope of this radius? The slope is 4/2sqrt(2) because we have a right-angled triangle made up of the y coord (-4), the x coord (-2sqrt(2)). We can ignore the negative sign, because we only need the magnitude of the sides. The hypotenuse is the radius and, indeed, 4^2+(2sqrt(2))^2=16+8=(2sqrt(6))^2=24. The slope 4/2sqrt(2)=sqrt(2). This is a positive slope. The tangent will have the slope -1/sqrt(2), normally written -sqrt(2)/2.
The equation of the tangent is y=-xsqrt(2)/2+b, where b is the y intercept. We know this line passes through (-2sqrt(2),-4), so we can find b. -4=2+b, so b=-6. Therefore, the equation of the tangent is y=-(xsqrt(2)/2+6).