Possibilities to get the amplitude of a signal at specific frequency from IQ
The easiest approach is to compute the one DFT bin of interest directly, which is a correlation to the frequency of interest by complex conjugate product and sum. Given you stated -10KHz frequency and with IQ samples then this frequency can be described using the exponential:

$$f_{in} = e^{-j\omega_n t} = I-jQ$$

Where 

$I = \cos(\omega_n t)$

$Q = \sin(\omega_n t)$

$\omega_n = 2\pi10 \text{ KHz}$

Expressing this in samples with $f_s = 250 \text{ KHz}$:

$I_n = \cos(\omega_n/f_s n) = \cos(2\pi n/25) $

$Q_n = \sin(\omega_n/f_s n) = \sin(2\pi n/25) $

The normalized correlation would give you the magnitude and phase:

$$X[\omega_n] = \frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{-j\omega_n} = \frac{1}{N}\sum_{n=0}^{N-1}x[n]\cos(2\pi n/25) + j\frac{1}{N}\sum_{n=0}^{N-1}x[n]\sin(2\pi n/25) $$

or equivalently in terms of the complex samples $I+jQ$:

$$X[\omega_n = -2\pi n/25] = \frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{j2\pi n/25}$$

Using an integer multiple of samples will result in an exact result as it would be the equivalent of a bin center for the DFT:

$$ N = m\frac{250 \text{ KHz} }{10 \text{ KHz} } = 25m$$

for any integer $m&gt;0$

The FFT performs this for all bins with $N\log_2(N)$ operations while here $N$ operations are given, thus when only one bin is needed computing that bin using the DFT directly results in a savings of $log_2(N)$. When multiple but still relatively small number of bins are needed, the Goertzel algorithm is a good choice.