What is the equation of a circle with a centre (6,-2) and passing through (2,-2)?Option: 1 $\left ( x-6 \right )^{2}+\left ( y+2 \right )^{2}= 16$Option: 2 $\left ( x-6 \right )^{2}+\left ( y+2 \right )^{2}= 36$Option: 3 $\left ( x-2 \right )^{2}+\left ( y+2 \right )^{2}= 16$Option: 4 $\left ( x-2 \right )^{2}+\left ( y+2 \right )^{2}= 36$

The equation of a circle with centre (h,k) and radius r is given by $\left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2}$

We are given that the centre of the circle is (6,-2) and it passes through (2,-2).

Here,
$h= 6$
$k= -2$

Using the distance formula, we can find the radius of the circle:

$r= distance\: between\left ( 6,-2 \right )and\left ( 2,-2 \right )$
$= \sqrt{\left ( 6-2 \right )^{2}+\left ( -2-2\left ( -2 \right ) \right )^{2}}$
$= \sqrt{\left ( 16+0 \right )}$
$= 4$

Therefore ,$r= 4$

Substituting the values of (h,k,r) into the equation of a circle, we get:
$\left ( x-6 \right )^{2}+\left ( y+2 \right )^{2}= 4^{2}$
$\left ( x-6 \right )^{2}+\left ( y+2 \right )^{2}= 16$

So, the correct answer is the option (a) $\left ( x-6 \right )^{2}+\left ( y+2 \right )^{2}= 16$