# Find the equation of the circle having (1, –2)

Question:

Find the equation of the circle having (1, –2) as its centre and passing through 3x + y = 14, 2x + 5y = 18

Solution:

Solving the given equations,

3x + y = 14 ……….1

2x + 5y = 18 …………..2

Multiplying the first equation by 5, we get

15x + 5y = 70…….3

2x + 5y = 18……..4

Subtract equation 4 from 3 we get

13 x = 52,

Therefore x = 4

Substituting x = 4, in equation 1, we get

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, we get

(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

Expanding the above equation we get

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.