Question

If the points A$(1,4)$ and B are symmetrical about the tangent to the circle $x^2+y^2-x+y=0$ at the origin then co-ordinates of B are

• A. $(1,2)$
• B. $(2,1)$
• C. $(4,1)$
• D. none of these

Answer 1

The correct option is C $(4,1)$

$(x-\frac{1}{2}{)}^2+(y+\frac{1}{2}{)}^2=\frac{1}{2}$

Ant tangent to this circle is given by

$(y+\frac{1}{2})=m(x-\frac{1}{2})+{\frac{1}{2}}^{\sqrt{m^2+1}}$

To find the tangent through origin we have to substitute it into the equation

$⟹\frac{1}{2}=\frac{-1}{2}m+\sqrt{\frac{m^2+1}{2}}$

$⟹(m+1{)}^2=2(m+1{)}^2$

$⟹1+m^2=2m$

$⟹(m-1{)}^2=0⟹m=1$

Tangent equation is

$2y+1=2x–1+2$

$⟹x=y$

Image of $A(1,4)$ about $x=y$ is $(4,1)$ and so $B=(4,1)$