Question

The line $y=x$ is a tangent at $(0,0)$ to a circle of radius is 1, then centre of the circle is

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

Answer 1

The correct option is C $(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

equatio of cicle with radius ${=}^{\prime }{1}^{\prime }$

$(x-a{)}^2+(y-b{)}^2=1$

$(a,b)\to center$

given $y=x$ is a target

slope of target $=1$

perpendicular distance from center to target radius target radius

$\Rightarrow \frac{|b-a|}{\sqrt{1^2+1^2}}=1$ equatio of target

$\Rightarrow \frac{|b-a|}{\sqrt{2}}=1$ $2(x-a)+2(y-b)y^{\prime }=0$

$\Rightarrow |b-a|=\sqrt{2}$ $\Rightarrow y^{\prime }=\frac{(x-a)}{(Y-B)}$

$Given{y}^{\prime }=1$ At $(x,y)=(0,0)$

$\Rightarrow 1=-\frac{a}{b}$

$\Rightarrow b=-a$

From (1) and (2)

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

$a=\frac{1}{\sqrt{2}}$ $△b=\frac{1}{\sqrt{2}}$

$b=\frac{-1}{\sqrt{2}}$ $(a,b)=(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

$\therefore (a,b)=(\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}})$