JEE Main 2022 — Circle Question with Solution

$$c_1:x^2+y^2=r^2$$ ; center = (0, 0) and radius = r

$$c_2:(x-1{)}^2+(y-1{)}^2=r^2$$ ; center = (1, 1) and radius = r

$$c_3:(x-2{)}^2+(y-1{)}^2=r^2$$ ; center = (2, 1) and radius = r

Distance of $$y=mx+c$$ line from center (0, 0) is,

$$\left|\frac{0+0+c}{\sqrt{m^2+1}}\right|=r$$ ..... (1)

Distance of $$y=mx+c$$ line from center (1, 1) is,

$$\left|\frac{m-1+c}{\sqrt{m^2+1}}\right|=r$$ ..... (2)

Distance of $$y=mx+c$$ line from center (2, 1) is,

$$\left|\frac{2m-1+c}{\sqrt{m^2+1}}\right|=r$$ .... (3)

From (1) and (2), we get

$$\left|\frac{c}{\sqrt{1+m^2}}\right|=\left|\frac{m-1+c}{\sqrt{1+m^2}}\right|$$

$$\Rightarrow m-1+c=\pm c$$ ..... (4)

taking positive sign,

$$m-1+c=c$$

$$\Rightarrow m-1=0$$

$$\Rightarrow m=1$$

From (2) and (3), we get

$$\left|\frac{m-1+c}{\sqrt{1+m^2}}\right|=\left|\frac{2m-1+c}{\sqrt{m^2+1}}\right|$$

$$\Rightarrow (m-1+c)=\pm (2m-1+c)$$ ...... (5)

taking positive sign,

$$m-1+c=2m-1+c$$

$$\Rightarrow m=0$$

By taking positive sign we get two different value of m so it is not acceptable.

From equation (4), taking negative sign,

$$m-1+c=-c$$

$$\Rightarrow m+2c-1=0$$ ..... (6)

From equation (5), taking negative sign

$$m-1+c=-(2m-1+c)$$

$$\Rightarrow 3m+2c-2=0$$ ..... (7)

Solving equation (6) and (7), we get

$$3m+1-m-2=0$$

$$\Rightarrow 2m=1$$

$$\Rightarrow m=\frac{1}{2}$$

$$\therefore$$ $$2c=1-\frac{1}{2}$$

$$\Rightarrow c=\frac{1}{4}$$

Putting value of $$m=\frac{1}{2}$$ and $$c=\frac{1}{4}$$ in equation (1), we get

$$r=\left|\frac{\frac{1}{4}}{\sqrt{1+\frac{1}{4}}}\right|$$

$$=\left|\frac{1}{4}\times \frac{2}{\sqrt{5}}\right|$$

$$=\frac{1}{2\sqrt{5}}$$

$$\therefore$$ $$20(r^2+c)$$

$$=20\left(\frac{1}{4\times 5}+\frac{1}{4}\right)$$

$$=20\left(\frac{1+5}{20}\right)$$

$$=6$$