If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154square units, then obtain the equation of the circle.

Since, diameters of a circle intersect at the centre of a circle,

2x – 3y = 5 -------- (i)

3x – 4y = 7 -------- (ii)

Solving the above mentioned equations,

6x – 9y = 15 [Multiplying equation (i) by 3}

6x – 8y = 14 [Multiplying equation (ii) by 2}

y = 1

y = -1

Putting y = -1, in equation (i)

2x – 3(-1) = 5

2x + 3 = 5

2x = 2

x = 1

Coordinates of centre = (1,-1)

Area = (Given)

r = 7 units

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

(x – 1)² + (y – (-1))² = 7²

x² - 2x +1 + (y + 1)² = 49

x² - 2x + 1 + y² + 2y + 1 – 49 = 0

x² - 2x + y² + 2y – 47 = 0

Hence the required equation of the given circle is x² - 2x + y² + 2y – 47 = 0.