Firstly it will help to draw a diagram of our two points along the x-axis, from this we can find out the x co-ordinate of the center of the circle as these points form a chord of the circle. Where in this case the x co-ordinate is the midpoint of two given points. So we have: (10+2)/2 = 6 which is our x co-ordinate for the center, if this is still unclear it may help to refer to the diagram. The y co-ordinate is still unknown for now. Next we know that as the y axis is a tangent line we have that the radius of the circle is 6 ( this may help to refer to the diagram drawn). Finally we need to find the y co-ordinate of our center, for this we can use our known equations for circles. We have: (x-6)2 + (y-a)2 =62 , as we know in order to obtain an unknown value we can substitute in a point we know the circle goes through, so at (2,0), x=2 and y=0, so 42 +a2 =36, re-arranging this gives a= sqrt(20) or a= -sqrt(20), so we have our two possible equations for the circle: (x-6)2+(y+sqrt(20))2=36 and (x-6)2 +(y-sqrt(20))2 =36