Question

A circle passes through the point $(6,2)$. If segments of the straight lines $x+y=6$ and $x+2y=4$ are two diameters of the circle, then its radius is

• A. $4$
• B. $8$
• C. $\sqrt{5}$
• D. $2\sqrt{5}$
• E. $4\sqrt{5}$

Answer 1

Answer: (D)

$2\sqrt{5}$

Since, the lines $x+y=6$ and $x+2y=4$ are two diameters of the circle, then their point of intersection will be the centre of circle.
Put $y=6-x$ in $x+2y=4$, we get
$x+2(6-x)=4$
$\Rightarrow x+12-2x=4$
$\Rightarrow -x=-8$
$\Rightarrow x=8$
Therefore, $y=6-8=-2$
Hence, centre of the circle is $(8,-2)$
Radius of circle $=$ Distance between $(8,-2)$ and $(6,2)$
$=\sqrt{{(8-6)}^2+{(-2-2)}^2}$
$=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$