CircleEasy
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.
Create a free account to view solution
View Solution FreeMore Circle Questions
The circle x2 + y2 - 2 x - 3 k y - 2 = 0 passes through two fixed points, (k is the parameter)...The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle of area 154 sq unit. Then the equation of this is...The equation of the circle which touches both the axes and the line + = 1 and lies in the first quadrant is (x _ c)2 + (...The intercepts made by the circle x2 + y2 _ 5x _ 13y _ 14 = 0 on the x-axis and y-axis are respectively...Consider four circles (x ± 1)2 + (y ± 1)2 = 1, then the equation of smaller circle touching these four circle ...