A circle with centre C(2, 3) passes through the point A(-4,-5). (a) Find the equation of the circle in the form (x-a)^2 + (y-b)^2=k

This question is aimed at A-Level Pure Core 1 students. A circle with centre C(2, 3) passes through the point A(-4,-5).  Find the equation of the circle in the form (x-a)^2 + (y-b)^2 = k From the formula for a circle above,  the centre coordinates are (a,b). We can substitute the information given above into the formula, replacing: a-->2 and b-->3 from the centre given in the question.  So far our answer looks like:  (x-2)^2 + (y-3)^2 = k Now we just have to work out the value of k. k represents the radius of the circle squared. To find the radius, we need to find the length of the line AC. To do this we can use Pythagoras' theorem (a^2 + b^2 = c^2). Making our line and also our radius, AC, form the hypotenuse (or the 'c' part of Pythagoras' theorem) of the right angled triangle with the axis lines, we can see the x - difference between A and C is 6 ( from 2 - -4 = 6) and the y - difference is 8 (from 3 - -5 = 8).  Therefore, using Pythagoras' theorem, we get 6^2 +8^2 = 36 + 64 = 100. Since k = radius squared = c^2 , we have our answer and can substitute this into the equation: (x-2)^2 + (y-3)^2 = 100