Find the equation of the circle having (1, –2) as its centre and pas

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

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

Answer»

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.