l1 and l2 are tangents of a circle. l1 intersects the circle at (3-√3,5) with a gradient of √3, and l2 intersects the circle at (3+√2,4+√2) with a gradient of -1. Find the centre of the circle, and hence find the radius of the circle.

To first find the centre of the circle, the key fact to remember is that the radius of a circle is normal (perpendicular) to any tangent. This means if you take the normals to both of these tangents at their respective intersection points, you will have two radii which will always intersect at the centre of the circle (by definition a radius is the length from the centre of a circle to its circumference). This means you will have to find the equations for the two normal normals of the two tangents, then find their intersection point. To find these you should use the fact that the gradient of a normal is the negative reciprocal of the tangent (1/gradient of tangent * -1), then substitute into y=mx+c to find their intersects. This will give you y=(-1/√3)x+(4+√3) and y=x+1 for l₁ and l₂ respectively. Set these equal to each other and you will get (3,4) as the centre of the circle.To find the radius use one of the intersects and the centre to find the change in y and change in x, then use Pythagoras's theorem to find the length of a radius, which will come out as 2.