Question

The co-ordinates of the point on the circle $x^2+y^2-12x-4y+30=0$ which is farthest from the origin are:

• A. $(9,3)$
• B. $(8,5)$
• C. $(12,4)$
• D. $noneofthese$

Answer 1

The correct option is A $(9,3)$

Given equation of circle $x^2+y^2-12x-4y+30=0$

$(x-6{)}^2-36+(y-2{)}^2-4+30=0(x-6{)}^2+(y-2{)}^2=10$

Maximum distance of $(0,0)$ from circle is

$=\sqrt{(6-0{)}^2+(2-0{)}^2}+\sqrt{10}=2\sqrt{10}+\sqrt{10}=3\sqrt{10}$

Let coordinates of point $P(x_1,y_1)$

then $\sqrt{(x_1-0{)}^2+(y_1-0{)}^2}=3\sqrt{10}\left(1\right)$

and this point also satisfies circle

$(x_1-6{)}^2+(y_1-2{)}^2=10\left(2\right)$

Solving $\left(1\right){x_1}^2+{y_1}^2=90\left(3\right)$

Solving $\left(2\right){x_1}^2+{y_1}^2-12x_1-4y_1+30=0\left(4\right)$

Solving $\left(3\right),\left(4\right)$

$12x_1+4y_1-30=9012x_1+4y_1=1203x_1+y_1=30\left(5\right)$

Equation of line passing through origin, center and point $(x_1,y_1)$ is

$(y-0)=\left(\frac{2}{6}\right)(x-0)3y=x$

Point $(x_1,y_1)$ also lies on this line

$3y_1-x_1=0\left(6\right)$

Solving $\left(5\right),\left(6\right)$

$3x_1+y_1=30$

$9y_1-3x_1=0$

$\overline{10y_1+0=30}$

$\Rightarrow y_1=3,x_1=9$

The point $(x_1,y_1)$ is $(9,3)$