Find the equation of the circle having (1, –2) as its centre and passing through3x + y = 14, 2x + 5y = 18

Solving the given equations,

3x + y = 14 ------- (1)

2x + 5y = 18 ------- (2)

Multiplying the first equation by 5,

15x + 5y = 70

2x + 5y = 18

13 x = 52, x = 4Putting x = 4, in first equation,

3(4) + y = 14

Y = 14 – 12 = 2

So, the point of intersection is (4,2)

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

(x – h)² + (y – k)² = r²

Putting the values of (4,2) and centre co-ordinates (1,-2) in the above expression,

(4 – 1)² + (2 – (-2))² = r²

3² + 4² = r²

r² = 9 + 16 = 25

r = 5 units

So, the expression is

(x – 1)² + (y – (-2))² = 5²

x² - 2x + 1 + (y + 2)² = 25

x² - 2x + 1 + y² + 4y + 4 = 25

x² - 2x + y² + 4y – 20 = 0

Hence the required expression is x² - 2x + y² + 4y – 20 = 0.