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