The equation of the circle with passes through the points (1, 0), (0, -6) and (3, 4) is
A. 4x2 + 4y2 + 142x + 47y + 140 = 0
B. 4x2 + 4y2 −142x − 47y + 138 = 0
C. 4x2 + 4y2 −142x + 47y + 138 = 0
D. 4x2 + 4y2 +150x − 49y + 138 = 0
Correct option C.
Explanation:
Equation of the required circle is:
x2 – x + y2 + 6y +24 + λ(6x – y – 6) = 0 →(1)
Put x = 3 and y = 4 in above equation:
λ = -23/4
Putting in (1) and rearranging:
4x2 + 4y2 – 142x + 47y + 138 = 0
