Question

Length of the chord $PQ$ of the circle $x^2+y^2-6x+8y-13=0$ whose midpoint is $(2,-3)$ is:

• A. $10$
• B. $11$
• C. $12$
• D. $13$

Answer 1

The correct option is C $12$

Equation of the circle is

$x^2+y^2-6x+8y-13=0$

$\Rightarrow x^2-6x+3^2-3^2+y^2+8y+4^2-4^2-13=0$

$\Rightarrow (x-3{)}^2+(y+4{)}^2-9-16-13=0$

$\Rightarrow (x-3{)}^2+(y+4{)}^3=38$

$\Rightarrow (x-3{)}^2+(y+4{)}^2=(\sqrt{38}{)}^2$

$\therefore$ Centre of the circle, $O=(3,-4)$

Radius, $r=\sqrt{38}$

Given midpoint of the chord PQ, $M=(2,-3)$

Draw $OM\perp PQ$

Distance/length of OM$=\sqrt{(3-2{)}^2+(-4+3{)}^2}$

$=\sqrt{1^2+(-1{)}^2}$

$=\sqrt{2}$

In right $\Delta$ OMP, using Pythagoras theorem

$O{P}^2=P{M}^2+O{M}^2$

$r^2=P{M}^2+(\sqrt{2}{)}^2$

$\Rightarrow P{M}^2=38-2=36$

$\Rightarrow PM=6$

In right $\Delta OMQ$, using Pythagoras theorem,

$O{Q}^2=O{M}^2+M{Q}^2$

$r^2=O{M}^2+M{Q}^2$

$M{Q}^2=38-2=36$

$\Rightarrow MQ=6$

$\therefore$ Length of $PQ=PM+MQ=6+6=12$.