Circles Part 1

Bonus or 4
x2+(y−β)2=r2
x−y=0
|

0−β

√2

|=r⇒β=r√12
x2+(y−β)2=

β2

2

⇒4−y+(y−β)2=

β2

2

⇒y2−y(2β+1)+

β2

2

+4=0
⇒(2β+1)2−4(

β2

2

+4)=0
4β2+4β+1−2β2−16=0
⇒2β2+4β−15=0
β=

−4±√16+120

4

=

−4±2√34

4

=

−2±√34

2

⇒

√34−2

2

r=

√34−2

2√2

Which is not in options therefore it must be bonus. But according to history of JEE-Mains it seems they had following line of thinking
Given curves are y=4−x2 and y=|x|
There are two circles satisfying the given conditions. The circle shown is of least area.
Let radius of circle is 'r
∴ co-ordinates of centre =(0,4–r)
∵circle touches the line y = x in first quadrant
∴|

0−(4−r)

√2

|=r⇒r−4=±r√2
∴r=4(√2−1)which is given in option 4.