Find the point on the circle which is nearest to the point (1, 2). <div style="float:left; width:4

Find the point on the circle $x^{2}+y^{2}= 80$ which is nearest to the point
(1, 2).

Answers (1)

Eqn of the circle is $x^{2}+y^{2}= 80$ and center is (0,0)

Let P(x,y) be the nearest point to A(1,2) which lies inside the circle
$OA= \sqrt{5}, OP= \sqrt{80}= 4\sqrt{5}$
$\therefore AP= OP-OA= 4\sqrt{5}-\sqrt{5}$
$= 3\sqrt{5}$
$OA:AP= \sqrt{5}:3\sqrt{5}$
$= 1:3$
The line containing the points (0,0) and (1,2) are y = 2x
Let the point on the circle be (h,k)
h = 2k, the point satisfies the equations
$h^{2}+k^{2}= 80$ the point lies on the circle
$\Rightarrow \left ( 2k \right )^{2}+k^{2}= 80$
$\Rightarrow \left ( 5k^{2} \right )= 80$
$\Rightarrow k^{2}= 16$
$\Rightarrow k= 4,-4$
h = 2k
$= 2\times 4= 8\: and\: 2\times -4= -8$
$\therefore$ The points that are possible are (4,8) and (-4,-8)