O-Level Equations and inequalities: worked solution

Question

The circle (x - 1)² + (y - 3)² = 5 is cut by the line y - 3x = 5 at a pair of points. Determine where these two points are.

One equation is linear and one is quadratic, so use substitution: make one letter the subject of the linear equation and put it into the other. The line is y − 3x = 5, so add 3x to both sides: y = 3x + 5 Substitute this expression for y into the circle (x − 1)² + (y − 3)² = 5: (x − 1)² + (3x + 5 − 3)² = 5 (x − 1)² + (3x + 2)² = 5 Expand each bracket separately: (x − 1)² = x² − 2x + 1 (3x + 2)² = 9x² + 12x + 4 Add them and keep the equation equal to 5: x² − 2x + 1 + 9x² + 12x + 4 = 5 10x² + 10x + 5 = 5 Subtract 5 from both sides: 10x² + 10x = 0 There is no constant term, so take out the common factor 10x rather than using the formula: 10x(x + 1) = 0 A product is zero when one of its factors is zero, so x = 0 or x + 1 = 0, i.e. x = −1 Substitute each x back into the line y = 3x + 5 to get its y: x = 0 gives y = 3(0) + 5 = 5 x = −1 gives y = 3(−1) + 5 = 2 The line cuts the circle at (0, 5) and (−1, 2).