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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.

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  • $\begingroup$ 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). $\endgroup$ – Aretino Sep 11 '15 at 16:34
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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?

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

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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.

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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.

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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

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