The circumference of a circle passing through the point $(4,6)$ with two normals represented by $2 x-3…
The circumference of a circle passing through the point $(4,6)$ with two normals represented by $2 x-3 y+4=0$ and $x+y-3=0$ is
$5 \pi$
$10 \pi$
$25 \pi$
$8 \pi$
Solution
Intersection of two normals will be the centre of the
circle.
$2 x-3 y+4=0$....(i) and $x+y-3=0$ ....(ii)
Solving equations (i) and (ii) we get, Centre $=(1,2)$
Also, circle passes through $(4,6)$
$\therefore$ Radius $(r)=\sqrt{(4-1)^2+(6-2)^2}=5$
So, circumference $=2 \pi r=10 \pi$.