Question

The locus of the point of intersection of the tangents at the ends of a chord of a circle $x^2+y^2=a^2$ which touches the circle $x^2+y^2-2ax=0$

• A. $y^2=a(a-2x)$
• B. $x^2=a(a-2y)$
• C. $x^2-y^2=(x-a{)}^2$
• D. $x^2+y^2=(y-a{)}^2$

Answer 1

The correct option is A $y^2=a(a-2x)$

Equation of chord to a circle $x^2+y^2+29n+2fy+c=0$ from $P(n_1,y_1)$ is given by $T=0$
$\therefore x{x}_1+y{y}_1+9(x+x_1)+f(y+y_1)+C=0$
Let $P(h,k)$ be point from which equation of chord to $x^2+y^2=a^2$ is
$xh+yk=a^2\left(1\right)$
But as $xh+yk-a^2=0$ is also a tangent to
$x^2+y^2-2an=0$ circle then $a=\left|\frac{ah+0-a^2}{\sqrt{(h^2+k^2)}}\right|$
$\sqrt{h^2+k^2}=|h-a|$
$\therefore h^2+k^2=(h-a{)}^2$
So, locus of P(h,k) is
$\Rightarrow x^2+y^2=(x-a{)}^2$