Question

Prove that the straight line $y=x+c\sqrt{2}$ touches the circle $x^2+y^2=c^2$, and find its point of contact.

Answer 1

If a line touches the circle, then distance between the centre and line is equal to the radius.

For the circle with centre $C(0,0)$ and $r=c$

Distance of centre from the line is

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

$d=\left|c\right|=r$

$c=c$

is true for all values of $c$

So, the line touches the circle

$y=x+c\sqrt{2}$ .....(1)

Putting $y$ in the equation of the circle

$x^2+(x+c\sqrt{2}{)}^2=c^2$

$x=\frac{-c}{\sqrt{2}},y=\frac{c}{\sqrt{2}}$

are the point of contact.