This one requires some circle properties which you have not learnt yet in E Maths.

Need my help? If yes, no one responds yet and I am able to spare the time at night, I will write down my workings.

The important symmetrical property of circles which you need to know is that the perpendicular bisector of any chord in a circle will pass through the centre of that circle. This can be proved by congruency and reasoning (I am not sure if you have covered congruency in E Maths yet).

A chord is simply a line joining two random points residing on the surface of the circle.

A perpendicular bisector to the chord means that you need to draw a line which is perpendicular to the chord and cuts the chord into two equal parts (ie passes through the midpoint of the chord or line segment).

So basically, you need to find the midpoint of (2, 3) and (-1, 6), which is going to be (0.5, 4.5). The perpendicular bisector will pass through this point.

You need to find the gradient of the line formed by (2, 3) and (-1, 6). This works out to be -1. The gradient of the bisector will then be 1, since the product of gradients of two perpendicular lines will be -1.

Then, the equation of the bisector will be

y = mx + c
y = x + c (since m = 1)

Using (0.5, 4.5),
4.5 = 0.5 + c
4 = c

So the centre of the circle will pass through the line y = x + 4.

Another line passing the same centre has been given in the question.

The two lines meet at the centre, so you will need to do simul on the two equations to find the values of x and y. This will be the coordinates of the centre of the circle.

To find the radius, you need to find the length between the centre and either of the two points (2, 3) or (-1, 6) on the circle.

You can then form the equation of the circle. I let you try on your own first.

If you still have difficulties, let me know.