Question

If $ax+by-3=0$ is the equation of the shortest chord of the circle ${(x-3)}^2+{(y-4)}^2=4$ passing through the point $(2,3)$, then $|a+b|$ is

• A. $\frac{6}{5}$
• B. $4$
• C. $1$
• D. $8$

Answer 1

The correct option is A $\frac{6}{5}$

The shortest chord through a point is always perpendicular to the line joining the center and the point i.e. the point will be the midpoint of the chord

Slope of line joining the center and point is

$\frac{4–3}{3–2}=1$

Slope of $ax+by–3=0$ is $-1$

$⟹\frac{-a}{b}=-1⟹a=b$

Also $(2,3)$ lies on the chord

$⟹2a+3b–3=0$

$⟹5a=3$

$⟹a=\frac{3}{5}=b$

$|a+b|=\frac{6}{5}$