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

The Correct Option is D

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