Do the circles with equations x^2 -2x + y^2 - 2y=7 and x^2 -10x + y^2 -8y=-37 touch and if so, in what way (tangent to each other? two point of intersection?)
We must first complete the square for both equations and get the equations in the (x-a)2+(y-b)2=r2 with the primary objective of determining the centres of both circles (a,b) . Through completing the square twice (once for x, once for y) for the equation of the first circle (x2 -2x + y2 - 2y = 7 ) we arrive at (x-1)2- (-1)2 +(y-1)2-(-1)2 =7. We then need to evaluate the constants (x-1)2-1+(y-1)2-1=7, and then collect all the constants on the right hand side (x-1)2+(y-1)2=7+1+1 which gives us (x-1)2+(y-1)2=9. This equation is now in the 'normal' equation of the circle form and we can now identify the centre1 of this circle as (1,1) and the radius1 is √9=3. (Remember to only look at the positive answer of √9 as the radius is a length and so it must be positive.)We must now complete the square in the same way for the equation of the second circle ( x^2 -10x + y^2 -8y=-37) as follows. (x-5)2-(-5)2+(y-4)2-(-4)2=-37 ... (x-5)2-25 +(y-4)2-16=-37 ... (x-5)2 +(y-4)2=-37 +25+16 ... (x-5)2 +(y-4)2=4. We now know that the centre of the second circle2 is (5,4) and the radius2 is √4=2. The penultimate step is to work out the distance between the centres of the circles (d). To do this you can visualise a right angled triangle with the hypotenuse as a line between the two centres. We can then determine the lengths of the sides of the triangle (5-1=4 and 4-1=3) and we can then use pythagorus to find the length of the hypotenuse which is also the length between the two centres (d). Using pythagorus (4)2+(3)2=d2 so 25=d2 and d=√25=5. Since the distance between the two centres (5) = the sum of the radii (3+2=5), the two circles are tangent to each other and one circle does not lie within the other.
