**Question:**

**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.**

**Solution:**

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

2x – 3y = 5 ………1

3x – 4y = 7 ………..2

Solving the above equations,

Multiplying equation 1 by 3 we get

6x – 9y = 15

Multiplying equation 2 by 2 we get

6x – 8y = 14

y = 1

y = -1

Putting y = -1, in equation 1, we get

2x – 3(-1) = 5

2x + 3 = 5

2x = 2

x = 1

Coordinates of centre = (1,-1)

Given area = 154

Area = πr2 = 154

22/7 × r2 = 154

r2 = 154 × 7/22

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.