Q. 85.0( 1 Vote )

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

Answer :

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.


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