The point on the circle x^2+y^2+6x+8y = 75 which is closest to the origin, is at what distance from the origin? (Taken from an MAT paper)
This is a good example of a question where intuition about geometry is important. Completing the square is a standard A level practice to find the equation of a circle. Doing this we get(x+3)2-9 +(y+4)2-16 = 75 Rearranging we find that (-3,-4) is the centre of the circle and the radius is 10. If we draw this diagram out, you can see using pythagoras that the distance of the centre to the origin is 5. The point on the circle that you get when you extend this line is also of distance 5 from the origin. Imagine moving that point a little bit away and seeing how the line between it and the origin would change. Hopefully, you can see that this distance would only become bigger and that the answer must be 5.However if I haven't explained this very well, read the following calculation carefully.Drawing a diagram, one can see that the point on the circle closest to the origin must be in the upper right hand side quadrant. We create a triangle with sides a,b,c.Side a - This distance we will call X and connects some random point P on the circle to the origin. Side b - This is the distance from P to the centre of the circle. Side c - This is the distance from the centre of the circle to the origin.Using prior knowledge, b = 5 and c = 10 We let the angle between a and c be called theta. Using the cos formula for triangles we get 100 = X2+ 25 - 10Xcos(theta). We rearrange this to get a function f of theta: f(theta) = X^2-10Xcos(theta) - 75. Differentiating and setting to 0 (with the aim of minimising this function for values of theta): d[f]/d[theta] = 10Xsin(theta) = 0. We then get theta = 0 or pi (radians). Using theta = 0 would give a maximum upon differentiating again so we know theta = pi gives us the necessary value.Substituting this value into f and setting it to 0 which we know already: X^2 + 10X -75 = 0. Solving this gives X = 5 which is the correct answer.
