Question

Point $A$ lies on the circle $x^2+y^2=10$ and its perpendicular distance from the line $L:2x+y=10$ is maximum. Then the image of this point about the line $L$ is

• A. $(8+2\sqrt{2},4+\sqrt{2})$
• B. $(4+2\sqrt{2},4+\sqrt{2})$
• C. $(4+\sqrt{2},8+2\sqrt{2})$
• D. $(-4-\sqrt{2},-8-2\sqrt{2})$

Answer 1

The correct option is A $(8+2\sqrt{2},4+\sqrt{2})$

Line perpendicular to the given line $L:2x+y=10$ is of the form -
$x-2y=c$
Now, we know that the points whose distance from the given line is maximum or minimum lie on the normal of the circle.
Thus the line $x-2y=c$ should pass through $(0,0)\Rightarrow c=0$

$\Rightarrow x=2y$
Using the above condition in the equation of the given circle, we get -
$4y^2+y^2=10\Rightarrow y=\pm \sqrt{2},x=\pm 2\sqrt{2}$

Farthest point according to the diagram would be
$(-2\sqrt{2},-\sqrt{2})$
Point of intersection of line $x=2y$ and $2x+y=10$ would be $(4,2)$, which is also the midpoint of the point $(-2\sqrt{2},-\sqrt{2})$ and its image $(x_1,y_1)$
$\Rightarrow (\frac{-2\sqrt{2}+x_1}{2},\frac{-\sqrt{2}+y_1}{2})=(4,2)$
$(x_1,y_1)=(8+2\sqrt{2},4+\sqrt{2})$