In order to answer this question, we must transform this equation for a circle into the general form of an equation of a circle; namely something like (x-a)2+(y-b)2=r2. Once we have done this, we know that the centre of the circle is at (a,b) and that the radius of the circle is r, so we can then draw it. So first let's collect some of the terms of our equation, so that we can easily complete the square. So, we get x2-20x+y2-24y+195=0. Now we will 'complete the square' twice to get the equation in the desired form. Recall that when we complete the square, we are taking an expression like x2+6x, say, and trying to put it into the form (x-k)2 +m for some k and m. The method is as follows: first take the coefficient of the linear term, halve it, add x to it, and then square this sum. So, in our example we get (x+3)2. Now, take the halved number (3, in our example) and square it and subtract it from this bracket. So, in our example we get (x+3)2-9, which, if we check, expands out to be x2+6x as required. With this in mind, we will complete the square twice to get (x-10)2-100+(y-12)2-144+195=0. Now we collect terms and simplify to see that the original equation is equivalent to (x-10)2+(y-12)2=49=72. So, from looking at the general circle equation, we see that this is a circle of radius 7 with centre at (10,12), which we can now draw.