Straight lines divide the circumference of the circle $x^2+y^2=100$ into two arcs whose lengths are in the ratio $3:1$ Find the equation of straight lines which pass through $(7,1)$,and divide the circumference of the circle $x^2+y^2=100$ into two arcs whose lengths are in the ratio $3:1$
My attempt:
As the required line is dividing the circumference in the ratio of 
$3:1$.Therefore,angle subtended by the required line on the center is $\frac{\pi}
{2}$
.But i could not find the equation of the lines.
I let the equation of line as $ax+by+c=0$ and it passes through $(7,1)$.So $7a+b+c=0$
Then i stuck.Please help me. 
 A: HINT.....Any line passing through $(7, 1)$ can be written as $$y-1=m(x-7)\rightarrow y-mx+7m-1=0$$
We require that the distance from the origin (the centre of the circle) to this line is $5\sqrt{2}$, so we can use the formula for the distance from a point to a line to set up an equation for $m$.
Can you take it from there?
A: HINT: 
Let the equation of the line be $y=mx+c$ passing through the point $(7, 1)$ then we have $$1=m(7)+c$$ $$7m+c=1\tag 1$$
Substituting $y=mx+c$ in the equation of circle $x^2+y^2=100$, we get $$x^2+(mx+c)^2=100$$ $$(1+m^2)x^2+2mc x+c^2-100=0\tag 2$$
Let, the roots of the above equation be $x_1$ & $x_2$ then 
$$x_1+x_2=-\frac{-2mc}{1+m^2}=\frac{2mc}{1+m^2}$$
$$x_1x_2=\frac{c^2-100}{1+m^2}$$
the points of intersection are $(x_1, y_1)$ & $(x_2, y_2)$
Now, the circumference $=2\pi\times 10=20\pi$ is divided in a ratio $3:1$ then the angle subtended by the small arc at the center $$=\frac{\text{arc length}}{\text{radius}}=\frac{5\pi}{10}=\frac{\pi}{2}$$  hence, the lines joining the points $(x_1, y_1)$ & $(x_2, y_2)$ to the center $(0, 0)$ will be normal to each other hence, we have $$m_1\times m_2=-1$$ 
$$\frac{y_1-0}{x_1-0}\times \frac{y_2-0}{x_2-0}=-1$$ 
$$x_1x_2+y_1y_2=0$$
$$x_1x_2+(mx_1+c)(mx_2+c)=0$$
$$(1+m^2)x_1x_2+2mc(x_1+x_2)+c^2=0$$ 
I hope you can take it from here to solve for the values of $m$ & $c$
A: And, by the way, check the distance from the origin (the centre of the circle) to the point (7, 1),- whether it >, < or = $5\sqrt{2}$ :)
Then write the equation of the line through the point (7, 1) and the origin,
and then the line perpendicular to it. Probably, this helps.
A: First its present equation is connecting ( 7,1) to (-1,7)  due to requirements of arc division, subtending  angle at origin should be $ 90^0,$ by rotation with $90^0$ angle.
$$ \dfrac{1-y}{7-x}=\dfrac{6}{-8} $$ 
Next, distance to origin is $5 \sqrt 2 $ ,so you have build a similar triangle $\sqrt 2 $ times zoomed with resp to origin, multiplying its intercepts or normal length from origin.
You can take the last step.
A: Let the slope if equation be-: $m$  the equation of line becomes 
$$mx-y=7m-1$$
$$(mx-y)/(7m-1)=1\tag1$$
Now homogenize the eq. $x2+y2=100$ by multiplying $100$ by square of eq. (1) $$ x2+y2=100\{(mx-y)/(7m-1)\}^2\tag2$$
Now, you can put the condition of $90^\circ$ angle subtended by the lines on centre that is $(a+b)=0$
Value of (a) and (b) can be calculated from from equation (2) ....
Hope it helps
