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 тАУ 2y = 3
a тАУ 2(a) = 3
-a = 3
a = тАУ 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 тАШrтАЩ is given by: (xтАУ p)2┬а+ (y тАУ q)2┬а= r2
Now by substitutingthe values in the equation, we get
(x тАУ (-3))2┬а+(y тАУ (-3))2┬а= 32
(x + 3)2┬а+(y + 3)2┬а= 9
x2┬а+6x + 9 + y2┬а+ 6y + 9 = 9
x2┬а+ y2┬а+6x + 6y + 9 = 0
тИ┤ The equation of thecircle is x2┬а+ y2┬а+ 6x + 6y + 9 = 0.