O-Level Coordinate geometry in two dimensions

Find A and B, the intersections of 4y = 2x + 1 and 3y - x = 4xy. From 4y = 2x + 1, x = (4y - 1)/2. Substituting: 3y - (4y-1)/2 = 4y(4y-1)/2, i.e. (2y+1)/2 = 8y² - 2y, giving 16y² - 6y - 1 = 0, so y = 1/2 or y = -1/8. Then A = (1/2, 1/2), B = (-3/4, -1/8). Gradient of AB = (1/2 - (-1/8)) / (1/2 - (-3/4)) = (5/8)/(5/4) = 1/2. Henry's line 2y - 4x = 5 is y = 2x + 5/2, gradient 2. Product of gradients = (1/2)(2) = 1, not -1. Since the product is not -1, the lines are not perpendicular, so Henry's claim does not hold.

From part (a) the circle has centre (2, −7) and radius² = 26. Start from the centre-radius form of a circle, (x − a)² + (y − b)² = r², where (a, b) is the centre. Substituting a = 2, b = −7 and r² = 26: (x − 2)² + (y − (−7))² = 26 (x − 2)² + (y + 7)² = 26 Now expand both brackets, since the required form has no brackets. (x − 2)² = x² − 4x + 4 (y + 7)² = y² + 14y + 49 So: x² − 4x + 4 + y² + 14y + 49 = 26 Collect the constants and move everything to one side: x² + y² − 4x + 14y + 53 − 26 = 0 x² + y² − 4x + 14y + 27 = 0 Compare this term by term with x² + y² + 2gx + 2fy + c = 0: 2g = −4, so g = −2 2f = 14, so f = 7 c = 27 Check using the standard results: centre = (−g, −f) = (2, −7), and r² = g² + f² − c = 4 + 49 − 27 = 26, both of which match part (a). x² + y² − 4x + 14y + 27 = 0, with g = −2, f = 7, c = 27

From the earlier parts of this question the circle has centre B(−4, −2) and radius √50, i.e. (x + 4)² + (y + 2)² = 50, and y = −x + 4 touches it at A(1, 3). Parallel lines have equal gradients, so the second tangent has gradient −1 and can be written y = −x + c. Key fact: the two tangents to a circle that share a gradient touch the circle at OPPOSITE ends of a diameter. So the second point of contact A' is the point diametrically opposite A, and the centre B is the midpoint of AA'. Let A' = (x', y'). Using the midpoint formula on A(1, 3) and A'(x', y') with midpoint B(−4, −2): (1 + x')/2 = −4 → 1 + x' = −8 → x' = −9 (3 + y')/2 = −2 → 3 + y' = −4 → y' = −7 So A' = (−9, −7). Now write the tangent through A'(−9, −7) with gradient −1, using y − y₁ = m(x − x₁): y − (−7) = −1(x − (−9)) y + 7 = −1(x + 9) y + 7 = −x − 9 y = −x − 16. Check by the perpendicular-distance method: for y = −x + c, i.e. x + y − c = 0, the distance from B(−4, −2) must equal the radius √50. |(−4) + (−2) − c| / √(1² + 1²) = √50 |−6 − c| = √50 × √2 = √100 = 10 so −6 − c = 10 or −6 − c = −10, giving c = −16 or c = 4. c = 4 is the tangent we were already given, so the other tangent is c = −16, i.e. y = −x − 16.