New Register here for RBI, SEBI & NABARD strategy workshop! April 9 at 7 PM

    Question

    Find the equation of the circle passing through the

    points (1, 2) and (2, -1), and whose center lies on the line x - 2y + 3 = 0.
    A (x - 2)^2 + (y + 1)^2 = 5 Correct Answer Incorrect Answer
    B (x + 1)^2 + (y - 2)^2 = 10 Correct Answer Incorrect Answer
    C (x - 3)^2 + (y + 1)^2 = 5 Correct Answer Incorrect Answer
    D (x - 1)^2 + (y + 2)^2 = 8 Correct Answer Incorrect Answer

    Solution

    Let the center of the circle be (h, k), and the equation of the circle be (x - h)^2 + (y - k)^2 = r^2. Since the circle passes through (1, 2) and (2, -1): (1 - h)^2 + (2 - k)^2 = r^2 and (2 - h)^2 + (-1 - k)^2 = r^2. Also, the center lies on the line x - 2y + 3 = 0, so h - 2k + 3 = 0. Solve this system of three equations to find h, k, and r. After solving: h = 1, k = -2, and r^2 = 8. The equation of the circle is (x - 1)^2 + (y + 2)^2 = 8.

    Practice Next