Question

The coordinates of the middle point of the chord $2x-5y+18=0$ cut off by the circle $x^2+y^2-6x+2y-54=0$ is:

• A. $(1,4)$
• B. $(2,4)$
• C. $(4,1)$
• D. $(1,1)$

Answer 1

The correct option is A $(1,4)$

Let the mid point of chord be $(h,k)$ and the centre of the circle as we can see from the equation of circle given is $(-g,-f)\Rightarrow (3,-1)$

In $△OMB$ and $△OMA$

$OB=OA+$radius

$MB=MA$ ($\because M$ is mid point)

$OM=OM$ (common)

$\Rightarrow △OMB\cong △OMA$ by s.s.s

$\Rightarrow \angle OMB=\angle OMA$ (cpct)

$\Rightarrow \angle OMB=\angle OMA={90}^0$

$\therefore MO$ is perpendicular to $AB$

$\therefore$ Product of slopes $=-1$

$\left(\frac{k+1}{h-3}\right)\times \left(\frac{2}{5}\right)=-1$

$\Rightarrow 2k+2=5(3-h)$

$\Rightarrow 2k+2=15-5h$

$\Rightarrow 5h+2k=13\to \left(1\right)$

$\because (h,k)$ lies o $2x-5y+18=0$

$\Rightarrow 2h-5k=-18\to \left(2\right)$

Multiplying both sides by $5$

$\Rightarrow 10h-25k=-90rightarrow\left(3\right)$

Multiplying both sides of $\left(1\right)$ by $\left(2\right)$

$\Rightarrow 10h+4k=26rightarrow\left(4\right)$

Comparing $\left(3\right)$ and $\left(4\right)$

$\Rightarrow 10h-25k=-90-(10h+4k=26)$

$-29k=-116$

$\Rightarrow k=4$

$\Rightarrow h=\frac{5k-18}{2}$

$=\frac{20-18}{2}=1$

$\therefore (1,4)$ Answer.