Co-ordinate Geometry A-level: The equation of a circle is x^2+y^2+6x-2y-10=0, find the centre and radius of the circle, the co-ordinates of point(s) where y=2x-3 meets the circle and hence state what we can deduce about the relationship between them.
We have that x2 + y2 + 6x - 2y - 10 = (x+3)2 + (y-1)2 - 20 = 0 (step was motivated by the equation of a general circle (x-a)2 + (y-b)2 = r2 )We know that the general circle has centre at (a,b) and the radius = rHence, comparing this with our equation we deduce that the centre is (-3, 1) and radius is sqrt(20).Next, we need to find the point where the line y = 2x - 3 intersects the circle. For them to intersect, they must have the same y value and x value at some point. Hence our y in the equation of the line should be the same as the y in the equation of the circle, so we can use the equation of the line in the circle to get:x2 + (2x-3)2 + 6x - 2(2x-3) - 10 = 0, expanding the brackets we get x2 + 4x2 - 12x + 9 + 6x - 4x + 6 - 10 = 0, simplifying this, we get:5x2 - 10x + 5 = 0, we can divide through by the common factor 5 to finally get our equation for our x value which is, x2-2x+1 = 0To solve this we factorise it as usual, which is (x-1)2 = 0, giving that our only solution is x = 1. We now know that y = 2x - 3, which implies for the value x = 1, y = 2(1) - 3 = -1, so (1,-1) is the only intersection point. If we were to look at a circle and the line (easier to see with a diagram), we can see that if the line only intersects the circle at one point, then we can conclude that it must be tangent to the circle otherwise it would pass into the circle and hence would have to cross another point if it wasn't.
