Question

The coordinates of the point on the circle $x^2+y^2-2x-4y-11=0$ farthest from the origin are

• A. $(2+\frac{8}{\sqrt{5}},1+\frac{4}{\sqrt{5}})$
• B. $(1+\frac{4}{\sqrt{5}},2+\frac{8}{\sqrt{5}})$
• C. $(1+\frac{8}{\sqrt{5}},2+\frac{4}{\sqrt{5}})$
• D. None of these

Answer 1

The correct option is B $(1+\frac{4}{\sqrt{5}},2+\frac{8}{\sqrt{5}})$

The equation of the given circle can be written as

${(x-1)}^2+{(y-2)}^2=16=4^2$

So, the coordinates of any point $P$ on the circle are $(1+4\cos \theta ,2+4\sin \theta )$ whose distance from the origin is

$d=\sqrt{{(1+4\cos \theta )}^2+{(2+4\sin \theta )}^2}=\sqrt{21+8\cos \theta +16\sin \theta }$

$\Rightarrow d=\sqrt{21+8(\cos \theta +2\sin \theta )}=\sqrt{21+8r\cos (\theta -\alpha )}$

where $r\cos \alpha =1,r\sin \alpha =2$

$=\sqrt{21+8\sqrt{5}\cos (\theta -\alpha )}$

which is maximum when $\cos (\theta -\alpha )=1.$

i.e., $\theta =\alpha ={\tan }^{-1}2.$

$\Rightarrow \tan \theta =2$ so $\sin \theta =\frac{2}{\sqrt{5}}$ and $\cos \theta =\frac{1}{\sqrt{5}}$

and, the coordinates of the required point are $(1+\frac{4}{\sqrt{5}},2+\frac{8}{\sqrt{5}})$