Find the equation of straight line which passes through the intersection of the lines $x-2y-5=0$ and $7x+y=50$ and divides the circumference of the circle $x^2+y^2=100$ into arcs of ratio 2:1
The point of intersection of the two lines was found to be $(7,1)$
Since the lines divides the arc in the ratio 2:1, the angles subtended by the line at the center is $2\pi/3$
If we consider $\Delta ABC$, where A is the origin, line BC is the required straight line and $D(h,k)$ is the foot of the perpendicular drawn from A.
Length AD is 5 units.
From here I can find the locus of D, but that isnt helpful. I just need the point D, and from there I can easily find the equation of the line.
Best Answer
Let $D(a,b)$. Then, the distance from the Origin to $D$ and the distance from $(7,1)$ to $D$ will give us 2 equations: $$ \begin{cases} a^2+b^2=25 \\ (a-7)^2+(b-1)^2=25 \end{cases} $$ By solving these, we get $(a,b)=(3,4)$ or $(4,-3)$
(Reason for the second equation: Let $M(7,1)$. Both $D$ and $M$ are located on the line $BC$. So, the triangle $ADM$ is a right triangle. $AM=5\sqrt{2}$ and the leg $AD=5$ implies that the leg $DM=5$)