Question -
Answer -
Let us assume thecircle touches the axes at (a, 0) and (0, a) and we get the radius to be |a|.
We get the centre ofthe 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 withcentre (-3, -3) and having radius 3.
We know that theequation of the circle with centre (p, q) and having radius ‘r’ is given by: (x– p)2 + (y – q)2 = r2
Now by substitutingthe values in the equation, we get
(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 thecircle is x2 + y2 + 6x + 6y + 9 = 0.