Question

If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 sq. units, then equation of the circle is (Taken π=$\frac {22}{7}$)

• A. x2 + y2 – 2x – 2y – 49 = 0
• B. x2 + y2 – 2x + 2y – 49 = 0
• C. x2 + y2 – 2x – 2y – 47 = 0
• D. x2 + y2 – 2x + 2y – 47 = 0

Answer 1

Answer: (D)

x2 + y2 – 2x + 2y – 47 = 0

Answer 2

Center is the point of intersection of diagonals.
2x−3y=5
3x−4y=7
Solving x=1,y=−1 Center =(1,−1) For radius r ( let )
r=$\sqrt {\frac {154}{π}}$
r=$\sqrt {\frac {154}{22\times 7}}$
r=7
(x−1)2+(y+1)2=72
​x2+1−2x+y2+1+2y=49
x2+y2−2x+2y−47=0
Therefore, the correct option is (D) x2+y2−2x+2y−47=0