The line 2x – y + 6 = 0 meets the circle x2 + y2 – 2y – 9 = 0 at A and B. Find the equation of the circle on AB as diameter.

Given that we need to find the equation of the circle whose diameter is AB.



It is also told that the points A and B are the intersection points when the line 2x - y + 6 = 0 meets the circle


x2 + y2 - 2y - 9 = 0. ..... - (1)


Let us first find the points A and B.


From the equation of the line:


2x - y + 6 = 0


y = 2x + 6 ..... (2)


Substituting (2) in (1), we get


x2 + (2x + 6)2 - 2(2x + 6) - 9 = 0


x2 + 4x2 + 24x + 36 - 4x - 12 - 9 = 0


5x2 + 20x + 15 = 0


5x2 + 5x + 15x + 15 = 0


5x(x + 1) + 15(x + 1) = 0


(5x + 15)(x + 1) = 0


5x + 15 = 0 or x + 1 = 0


5x = - 15 or x = - 1


x = - 3 or x = - 1


For x = - 3, from (2)


y = 2(- 3) + 6


y = - 6 + 6


y = 0


For x = - 1, from (2)


y = 2(- 1) + 6


y = - 2 + 6


y = 4


The points are A(- 3,0) and B(- 1,4)


We know that centre is the mid - point of the diameter.


Centre(C) =


C = (- 2,2)


We have circle with centre (- 2,2) and passing through the point (- 1,4).


We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.


We know that the distance between the two points (x1,y1) and (x2,y2) is .


Let us assume r is the radius of the circle.





r = √5


We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:


(x - p)2 + (y - q)2 = r2


Now we substitute the corresponding values in the equation:



x2 + 4x + 4 + y2 - 4y + 4 = 5


x2 + y2 + 4x - 4y + 3 = 0


The equation of the circle is x2 + y2 + 4x - 4y + 3 = 0.


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