Question

The value of $c$ for which $x+y=2$ touches circle $x^2+y^2-2cx=0$ and $(1,0)$ lies inside the circle, is $2$

• A. $2$
• B. $\sqrt{2}-1$
• C. $2(\sqrt{2}-1)$
• D. $2(\sqrt{2}+1)$

Answer 1

The correct option is D $2(\sqrt{2}+1)$

REF.Image.

$x+y=2$

$x^2+y^2-2cx=0$

$c=(c,0),8=c$

if this line touch

then if will be tangent

$c^2=a^2(1+m^2)$

${\perp }^{rl}$ distance will be it's Radius

$\left|\frac{c+0-2}{\sqrt{2}}\right|=c$

$\frac{c-2}{\sqrt{2}}=c$

$c-2=\sqrt{2c}$

$c(1-\sqrt{2})=2$

$c=\frac{2}{1-\sqrt{2}}\frac{(1+\sqrt{2})}{(1+\sqrt{2})}$

$c=\frac{2(1+\sqrt{2})}{1}$