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Question: The line joining (5, 0) to (10\(\cos \theta \), 10\(\sin \theta \)) is divided internally in the rat...

The line joining (5, 0) to (10cos⁡θ\cos \theta , 10sin⁡θ\sin \theta ) is divided internally in the ratio 2:3 at P. If θ\theta varies, then the locus of P is:
(a) A pair of straight lines
(b) A circle
(c) A straight line
(d) None of these

Explanation

Solution

Hint:Let coordinates of P be (x,y)\left( x,y \right). Use section formula to determine the coordinates of P. section formula is given by (x,y)\left( x,y \right)=(mx2+nx1m+n,my2+ny1m+n)\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right), where m and n are the ratio in which the line segment is divided internally. Then establish the relationship between x and yx\text{ and }y.

Complete step-by-step answer:
Let us assume, (x1,y1)({{x}_{1}},{{y}_{1}})= (5, 0), and (x2,y2)({{x}_{2}},{{y}_{2}})= (10cos⁡θ\cos \theta , 10sin⁡θ\sin \theta ).Then, mm must be 2 and nn must be 3.

Now, let us calculate the coordinates of P.

& x=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n} \right) \\\ & \text{ }=\left( \dfrac{2\times 10\cos \theta +3\times 5}{2+3} \right) \\\ & \text{ }=\left( \dfrac{20\cos \theta +15}{5} \right) \\\ & \text{ }=\left( 4\cos \theta +3 \right) \\\ \end{aligned}$$ Now, $\begin{aligned} & y=\left( \dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right) \\\ & \text{ }=\left( \dfrac{2\times 10\sin \theta +3\times 0}{2+3} \right) \\\ & \text{ }=\left( \dfrac{20\sin \theta }{5} \right) \\\ & \text{ }=4\sin \theta \\\ \end{aligned}$ Hence, P$\left( x,y \right)$=$\left( 4\cos \theta +3,4\sin \theta \right)$ Now, $\begin{aligned} & x=4\cos \theta +3 \\\ & \therefore (x-3)=4\cos \theta ....................................(i) \\\ \end{aligned}$ Also, $y=4\sin \theta .......................................(ii)$ On squaring and adding equation (i) and (ii), we get, $$\begin{aligned} & {{(x-3)}^{2}}+{{y}^{2}}=16{{\sin }^{2}}\theta +16{{\cos }^{2}}\theta \\\ & {{(x-3)}^{2}}+{{y}^{2}}=16({{\sin }^{2}}\theta +{{\cos }^{2}}\theta ) \\\ & {{(x-3)}^{2}}+{{y}^{2}}=16 \\\ & {{(x-3)}^{2}}+{{y}^{2}}={{4}^{2}} \\\ \end{aligned}$$ This is of the form ${{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}$, which is an equation of the circle with centre (a,b) and radius r. Hence, option (b) is the correct answer. Note: One may get confused in selecting the value of m and n. We don’t have to worry about that. Here, select any one of them as m and n but, remember that the side towards which we have selected ‘m’ must be denoted as coordinate $({{x}_{1}},{{y}_{1}})$ and the other side which is selected as ‘n’ must be denoted as coordinate$({{x}_{2}},{{y}_{2}})$. Then apply the section formula.