Q. 94.0( 9 Votes )

Find the equation of the circle which touches the axes and whose centre lies on x – 2y = 3.

Answer :

We need to find the equation of the circle the axes and centre lies on x - 2y = 3.



Let us assume the circle touches the axes at (a,0) and (0,a) and we get the radius to be |a|.


We get the centre of the circle as (a, a). This point lies on the line x – 2y = 3


a - 2(a) = 3


- a = 3


a = - 3


Centre = (a, a) = ( - 3, - 3) and radius of the circle(r) = | - 3| = 3


We have circle with centre ( - 3, - 3) and having radius 3.


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:


(x - ( - 3))2 + (y - ( - 3))2 = 32


(x + 3)2 + (y + 3)2 = 9


x2 + 6x + 9 + y2 + 6y + 9 = 9


x2 + y2 + 6x + 6y + 9 = 0.


The equation of the circle is x2 + y2 + 6x + 6y + 9 = 0.


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