Question

Locus of the mid point of the chord of circle $x^2+y^2=16$ which is subtending right angle at the point (5, 0) is

• A. $x^2+y^2+4x+5/2=0$
• B. $x^2+y^2-5x+9/2=0$
• C. $x^2+y^2+5x+7=0$
• D. $x^2+y^2+2x+5=0$

Answer 1

The correct option is B $x^2+y^2-5x+9/2=0$

Let the mid point be (h,k)

$T=S_1$ for the equation of chord , the chord is $hx+ky=h^2{k}^2$

If the chord intersects the circle at$\left(x_1{y}_1\right),(x_2,y_2)$,then by condition of perpendicularity

$m_1{m}_2=-1\Rightarrow \left(\frac{y_1-\beta }{x_1-\alpha }\right)\left(\frac{y_2-\beta }{x_2-\alpha }\right)=-1$

here the given circle is of the form $x^2+y^2=a^2$.The tangents intersect at some point say $(\alpha ,\beta )$

$(x_1{x}_2+y_1{y}_2)+({\alpha }^2+{\beta }^2)=\alpha (x_1+x_2)+\beta (y_1+y_2)\lambda x^2-2\lambda hx+{\lambda }^2-a^2{k}^2=0\lambda y^2-2\lambda ky+{\lambda }^2-a^2{h}^2=0[\because \lambda ={\lambda }^2+k^2]\Rightarrow x^2+y^2-\alpha x-\beta y+\frac{1}{2}({\alpha }^2+{\beta }^2-a^2)=0$

where $(\alpha ,\beta )$ ,(5,0) and $x^2+y^2=16$ where a=4

$\therefore x^2+y^2-5x+\frac{1}{2}[25-16]=0\Rightarrow x^2+y^2-5x+\frac{9}{2}=0$