Show that the equation of the circle which touches the coordinates axes whose centre lies on the line `l x+m y+n=0\ i s\ (l+m)^2 (x^2 + y^2 )+2n(x+y)(l+m)+n^2=0.`

Explanation

We know that coordinates of the centre of a circle toruching the coordinates axes in first quadrant are` (a, a),` while` a` is the radius of the circle. So, the equation of the circle is –


`(x-a)^{2}+(y-a)^{2}=a^{2} or, x^{2}+y^{2}-2 a x-2 a y+a^{2}=0`


Since the centre `(a, a) `lies on `l x+m y+n=0.`


Therefore.


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