Question

A circle touches axesat (2,0)and also the line y=x in first quadrant then its radius is

Answer 1

Answer:

Radius (r)= perpendicular distance on line 4x+3y=12 from centre

⇒ r=r=16+9∣4r+3r−12∣

⇒∣7r−12∣=5r

⇒7r−12=±5r

∴2r=12⇒r=6

and 12r=12

⇒r=1

(i) When centre is (1,1) and radius is 1, then equation of circle is

(x−1)2+(y−1)2=1

⇒x2+y2−2x−2y+1=0

(ii) When centre is (6,6) and radius is 2, then equation of circle is

(x−6)2+(y−6)2=36

⇒x2+y2−12x−12y+36=0