The points (5, 5), (6, 4), (- 2, 4) and (7, 1) all lie on a circle. circle passes through the points A, B, C. therefore, the equation of the circle: x2 + y2 + 2ax + 2by + c = 0….. (1) Substituting A...
The points (3, 7), (5, 5) The line x – 4y = 1…. (1) the equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (2) substituting the centre (-a, -b) in equation (1) we get,...
The line 2x – y = 3 … (1) The points (3, -2), (-2, 0) The equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (2) substituting the centre (-a, -b) in equation (1) we get,...
(iii) equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (1) Substituting the point (5, -8) in equation (1), we get \[\begin{array}{*{35}{l}} {{5}^{2}}~+\text{ }{{\left( -\text{ }8...
(i) (5, 7), (8, 1) and (1, 3) By using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (1) Substitute the given point (5, 7) in equation (1), we get...
The equation of the circle is (Multiply by 2 we get) \[\begin{array}{*{35}{l}} {{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }2x\text{ }cos\text{ }\theta \text{ }+\text{ }2y\text{ }sin\text{ }\theta \text{...
(i) The equation of the circle is x2 + y2 + 6x – 8y – 24 = 0 …… (1) Since, for a circle x2 + y2 + 2ax + 2by + c = 0 …… (2) Centre = (-a, -b) So by comparing equation (1) and (2) Centre =...