# 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 $7a+b+c=0$

• First of all you can set $a=1$ and eliminate $c$, so that your line equation depends on $b$ only. Then you can find the points of intersections $A$ and $B$ between line and circle and fix $b$ so that $AB^2=10^2+10^2$ (Pythagoras' theorem). – Aretino Sep 11 '15 at 16:34

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?

• Sir,@David Quinn,there are two lines $x-2y-5=0$ and $7x+y-50=0$ given in the answer.But using this method,i am getting only $7x+y-50=0$. – Vinod Kumar Punia Sep 12 '15 at 2:24
• You should get a quadratic equation in m and hence two answers – David Quinn Sep 12 '15 at 3:56
• The quadratic equation $m^2+14m+49=0$ is giving me only one value of $m$ @David Quinn – Vinod Kumar Punia Sep 12 '15 at 3:59
• Are you sure the other answer is correct? The distance from the origin to it is $\sqrt{5}$ – David Quinn Sep 12 '15 at 9:22

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 $m$ & $c$

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.

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.

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