If the line 2x – y + 1 = 0 touches the circle at the point (2, 5) and the centre of the circle lies on the line x + y – 9 = 0. Find the equation of the circle.

Given that the line 2x - y + 1 = 0 touches the circle at the point (2,5) and the centre lies on the line x + y - 9 = 0.



We know that normal at any point on the circle passes through the centre of the circle.


We have tangent 2x - y + 1 = 0 at the point (2,5) for the circle.


We know that normal is perpendicular to the tangent and passes through the point of contact.


We know that product of slopes of two perpendicular lines is - 1.


Let us assume m be the slope of the normal.


We know that slope of a line ax + by + c = 0 is .


Slope of the line 2x - y + 1 = 0 is


m.(2) = - 1


- - - - - (1)


We know that the equation of a straight line passing through the point (x1, y1) and having slope ‘m’ is given by (y - y1) = m(x - x1).



2y - 10 = - x + 2


x + 2y - 12 = 0 ..... (2)


We get the centre by solving for the intersecting point of the lines x + 2y - 12 = 0 and x + y - 9 = 0 as they both pass through the centre.


On solving these two lines we get,


Centre = (6, 3)


We have circle with centre (6, 3) and passing through the point (2, 5).


We know that 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.





..... (2)


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 - 12x + 36 + y2 - 6y + 9 = 20


x2 + y2 - 12x - 6y + 25 = 0.


The equation of the circle is x2 + y2 - 12x - 6y + 25 = 0.


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