The centre of a circle is $(4a-2,6a+2)$ and is passing through the point $(-6,-2)$.  If the diameter of the circle is 40, then $a$ is equal to 

Given a circle with centre $(4a-2,6a+2)$ passing through the point $(-6,-2)$

Diameter $=40$

Radius $=20$

$\begin{array}{l}\therefore 20=\sqrt{(4a-2+6)+{(60+2+2)}^2}\\ \Rightarrow 20=\sqrt{{(4a+4)}^2+{(60+4)}^2}\\ \Rightarrow 400=16a^2+16+32a+36a^2+16+48a\\ \Rightarrow 13a^2+20a-92=0\\ \therefore a=\frac{-20\pm \sqrt{400-4\times -92\times 13}}{26}\\ =\frac{-20\pm \sqrt{400+4784}}{26}\\ =\frac{-20+72}{26}[\because -\text{ve\hspace{0.17em}\hspace{0.17em}sign\hspace{0.17em}\hspace{0.17em}neglected}]\\ =2\end{array}$