Question

The mid point of the chord $x-2y+7=0$ w.r.t the circle $x^2+y^2-2x-10y+1=0$ is?

• A. $(7,21)$
• B. $(\frac{7}{4},\frac{21}{4})$
• C. $(\frac{7}{5},\frac{21}{5})$
• D. $(\frac{-7}{5},\frac{-21}{5})$

Answer 1

The correct option is C $(\frac{7}{5},\frac{21}{5})$

$\text{Equation of circle}=x^2+y^2-2x-10y+1=0$

$\text{So;}\mathrm{center}=(1,5)$

$\text{Equation of chord}=x-2y+7=0$

$P=$ mid-point of chord $=(h,k)$

Now, $P$ must satisfy the equation of chord;

So;

$x-2y+7=0$

$h-2k+7=0$ - (1)

Now; slope of $AB$ $X$ slope of $PC=-1$

slope of $AB=\frac{1}{2}$ (given equation)

$\text{So;}\frac{1}{2}\times \left(\frac{k-5}{h-1}\right)=-1$

$2h+k-7=0$ - (2)

Solve equation (1) and (2)

$h=7/5\text{and}k=\frac{21}{5}$