Question

Match the entries of List-A and List-B.
List-A (Locus) List-B (Equations)

Answer 1

[A].

The equation of tangent to the circle $x^2+y^2=a^2$ is

$y=mx+a(1+m^2)$

$P(h,k)$ lies on tangent then

$(k-mh)=a1+m^2$

$(k-mh{)}^2=a^2(1+m^2)$

$m^2(h^2-a^2)-2mhk+k^2-a^2=0$

This is the quadratic equation in $m$,let two roots $m_1$ and $m_2$

$m_1{m}_2=-1$(⊥ tangents)

$k^2-a^2{h}^2-a^2=-1$

$k^2-a^2=-h^2+a^2$

$h^2+k^2=2a^2$

Hence locus $P(h,k)$ is $x^2+y^2=2a^2$

[B].

Let locus be $(h,k)$, then

$h=2+4cos\theta$ and $k=-1+4sin\theta$

$⟹(h-2{)}^2+(k+1{)}^2=(4cos\theta {)}^2+(4sin\theta {)}^2$

$⟹(x-2{)}^2+(y+1{)}^2=16$

[C].

Let locus be $(h,K)$ and points area$(x_1,y_1),(x_2,y_2)$. Then

$(x_1-h{)}^2+(y_1-k{)}^2=(x_2-h{)}^2+(y_2-k{)}^2$

$x_1^2-2h{x}_1+y_1^2-2k{y}_1=x_2^2-2h{x}_2+y_2^2-2k{y}_2$

$k=-\frac{x_2-x_1}{y_2-y_1}h$

$y=-\frac{1}{m}x$....which is equation of perpendicular bisector.

[D].

Locus of mid-points of the chord passing through $(x_1,y_1)$ of a circle (or conic) $x^2+y^2=a^2$is given by:

$T=S_1$

$x^2+y^2-a^2=x{x}_1+y{y}_1-a^2$

$x^2+y^2-x{x}_1-y{y}_1=0$