Find the equation of the circle passing through the points (1,-5).(5,7) and (-5,1).
Answers
Step-by-step explanation:
By using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (1) Firstly let us find the values of a, b and c Substitute the given point (5, 7) in equation (1), we get 52 + 72 + 2a (5) + 2b (7) + c = 0 25 + 49 + 10a + 14b + c = 0 10a + 14b + c + 74 = 0….. (2) By substituting the given point (8, 1) in equation (1), we get 82 + 12 + 2a (8) + 2b (1) + c = 0 64 + 1 + 16a + 2b + c = 0 16a + 2b + c + 65 = 0….. (3) Substituting the point (1, 3) in equation (1), we get 12 + 32 + 2a (1) + 2b (3) + c = 0 1 + 9 + 2a + 6b + c = 0 2a + 6b + c + 10 = 0….. (4) Now by simplifying the equations (2), (3), (4) we get the values a = -29/6, b = -19/6, c = 56/3 Substituting the values of a, b, c in equation (1), we get x2 + y2 + 2 (-29/6)x + 2 (-19/6) + 56/3 = 0 x2 + y2 – 29x/3 – 19y/3 + 56/3 = 0 3x2 + 3y2 – 29x – 19y + 56 = 0 ∴ The equation of the circle is 3x2 + 3y2 – 29x – 19y + 56 = 0Read more on Sarthaks.com - https://www.sarthaks.com/805517/find-the-equation-of-the-circle-passing-through-the-points-i-5-7-8-1-and-1-3-ii-1-2-3-4-and-5-6