Question

If $ax+by-5=0$ 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. $2$
• B. $4$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

$2$

$\because$ Perpendicular is the shortest distance from any point to a line and the perpendicular from the circle's center will always bisect the cord

$\therefore$ The shortest chord through a point is always perpendicular to the line(AC) joining the center and the point i.e. the point will be the midpoint of the chord

putting value of $(2,3)$ in the equation of the circle we get that,
$(2-3{)}^2+(3-4{)}^2-4=2-4=-2<0$
Hence, the point lies inside the circle

Slope of line joining the center and point is

$\frac{4-3}{3-2}=1$

Slope of $ax+by-5=0$ is $-1$

$\Rightarrow -\frac{a}{b}=-1$

$\Rightarrow a=b$

Also, $(2,3)$ lies on the chord $ax+by-5=0$

$\Rightarrow 2a+3b-5=0$

$\Rightarrow 5a-5=0$ and $5b-5=0$ since $a=b$

$\Rightarrow a=b=1$

$|a+b|=1+1=2$