Question

The normal to the circle given by $x^2+y^2-6x+8y-144=0$ at $(8,8)$ meets the circle again at the point

• A. $(2,-16)$
• B. $(2,16)$
• C. $(-2,16)$
• D. $(-2,-16)$

Answer 1

The correct option is D $(-2,-16)$

The equation of the circle can be put in center-radius form as:

$x^2+y^2-6x+8y-144=0\Rightarrow (x-3{)}^2+(y+4{)}^2=169$

So, clearly the radius of the circle is $13$ and its center is at $(3,-4)$.

So, let the normal at $(8,8)$ meets the circle again at $(x_0,y_0)$, then the midpoint of $(8,8)$ and $(x_0,y_0)$ is the center $(3,-4)$. Thus,

$x_0=6-8=-2$ and $y_0=-8-8=-16$

So the required point is $(-2,-16)$. Hence option D is the right answer.