Question

If a circle with the point $(-1,1)$ as its center touches the straight line $x+2y+9=0$ then the coordinates of the points of contact is

• A. $(-3,3)$
• B. $(-3,-3)$
• C. $(0,0)$
• D. $(\frac{7}{3},-\frac{17}{3})$

Answer 1

Answer: (B)

$(-3,-3)$

The normal to the tangent will pass through the centre. Also, the intersection of the normal and tangent will give the point of contact.
Hence consider the equation of the tangent.
$y=\frac{-x}{2}-4.5$
Hence $m=\frac{-1}{2}$.

Therefore the slope of the normal is $2$ since slope of normal and tangent are mutually perpendicular.
Therefore equation of the normal at $(1,-1)$ is
$\frac{y-1}{x+1}=2$ or $y-1=2x+2$ or $2x-y=-3$.

Therefore, we have two equations
$2x-y=-3$ and $x+2y+9=0$.

Solving these two equations give us
$(x,y)=(-3,-3)$.
Hence, the point of contact is $(-3,-3)$.