CircleHard
Question
If the lines 3x - 4y - 7 = 0 and 2x - 3y - 5 = 0 are two diameters of a circle of area49π square units, the equation of the circle is
Options
A.x2 + y2 + 2x - 2y - 47 = 0
B.x2 + y2 + 2x - 2y - 62 = 0
C.x2 + y2 - 2x + 2y - 62 = 0
D.x2 + y2 - 2x + 2y - 47 = 0
Solution
Point of intersection of 3x - 4y - 7 = 0 and 2x - 3y - 5 = 0 is (1 , - 1), which is the centre of the circle and radius = 7.
∴ Equation is (x - 1)2 + (y + 1)2 = 49 ⇒ x2 + y2 - 2x + 2y - 47 = 0.
∴ Equation is (x - 1)2 + (y + 1)2 = 49 ⇒ x2 + y2 - 2x + 2y - 47 = 0.
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