Question

The normal to the circle $x^2+y^2-6x+8y-144=0$ 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 normal of a circle passes through the centre of the circle and intersects the circle again at the diametrically opposite point. Let the other point be $A\equiv (h,k)$.

The centre of the circle is $(3,-4)$ and one end of the diameter is $(8,8)$. Since the centre is the midpoint of the diameter, we can say

$(3,-4)\equiv (\frac{8+h}{2},\frac{8+k}{2})$

$\Rightarrow h=2\left(3\right)-8$ and $k=2(-4)-8$

$\Rightarrow h=-2$ and $k=-16$

Hence the point is $A\equiv (-2,-16)$