Straight lines divide the circumference of the circle $x^2+y^2=100$ into two arcs whose lengths are in the ratio $3:1$

Question

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.

Answer 1

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?

Answer 2

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 $m$ & $c$

Answer 3

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.

Answer 4

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.