JEE Main 2020 — Mathematics Coordinate Geometry

Centre of circle lies on the line $x+y=2$x+y=2.

Any point on the line can be assumed as $(\alpha ,2-\alpha )$(α,2−α).

So let Centre of circle is $(\alpha ,2-\alpha )$(α,2−α).

Since both lines $x=3$x=3 and $y=2$y=2 touches the circle means distance from the Centre of circle to the line is equal to radius.

Let $r$r is radius of circle.

For the line $x=3\Rightarrow x-3=0$x=3⇒x−3=0

$$r=|\frac{\alpha -3}{1}|=|\alpha -3|...(1)$$r=∣1α−3​∣=∣α−3∣...(1)

For the line $y=2\Rightarrow y-2=0$y=2⇒y−2=0

$$r=|\frac{(2-\alpha )-2}{1}|=|\alpha |...(2)$$r=∣1(2−α)−2​∣=∣α∣...(2)

From $(1)$(1) and $(2)$(2)

$$|\alpha |=|\alpha -3|$$∣α∣=∣α−3∣

Squaring both side

$$\Rightarrow {\alpha }^{2}={(\alpha -3)}^{2}\Rightarrow {\alpha }^{2}={\alpha }^{2}-6\alpha +9$$⇒α2=(α−3)2⇒α2=α2−6α+9

$$\Rightarrow 6\alpha =9\Rightarrow \alpha =\frac{3}{2}$$⇒6α=9⇒α=23​

We know $r=|\alpha |\Rightarrow r=\frac{3}{2}$r=∣α∣⇒r=23​.

Diameter$=2r=2(\frac{3}{2})=3$=2r=2(23​)=3.