Adding offset before second order active filter
try







simulate this circuit &ndash; Schematic created using CircuitLab

You've rather got in a mess with which resistors are biasing and which doing signal. 

Now I've finished my tea, and thought a bit more, and read the comments ...

R1-R2 provides mid rail bias of the op amp, and the right impedance to ground for the filter network.

It looks like the intention of your R9,10 was to provide the bias, but then you've connected them in parallel to your inverting input. What you need is a network of two resistors that simulataneously provides the right voltage, and the right impedance to ground. My R1,2 provide mid rail, but it would be equally as easy to use for instance 1.6k and 4.7k to bias to 0.25x rail with 1.2k impedance to ground.

C3 is a better more generic solution for getting the DC offset into the feedback path, which in this case of a high pass filter, is obviously the case. It will still work if you ask for gain of 1000, which separate biassing probably won't. If correct gain and offset is important down to DC, then R5 should be replaced by a 2k resistive divider like R1R2.

Your R7 looks like it's part of a Sallen Key filter, but unfortunately the gain is wrong. With an equal component SK, a gain of unity gives you quite low Q. As you increase the gain, the Q increases, becoming infinite at a gain of 3. In fact such an equal component gain 3 SK filter is called a Wein Bridge Oscillator! Gains above 3 just give you instability and crashing into the rails.

R3,7 may be the clever bit. It undoes the x10 gain of the opamp, and presents the signal to the caps' mid point at the impedance of R3//R7, the 1.2k in your original. This will give you back the response of a unity gain equal component sallen key, followed by x10 gain, with one opamp. 

However, while such a filter is stable, high pass, and has ultimate rejection at 40dB/decade, the passband is not one of the recognised responses. This may or may not matter. You could increase the gain at the filter by reducing R3 and increasing R7 a little, while keeping their parallel impedance 1.2k, to choose responses in the Butterworth or other families.