Question

Locus of P.O.I of tangents to $x^2+y^2=4$. If corresponding chord of contact touch $x^2+y^2=1$ is given as $x{x}_1+y{y}_1=1$

• A. (a)$x^2+y^2=1$
• B. (b)$3x-y=1$
• C. (c)$x^2+y^2=16$
• D. (d)$3x+y=3$

Answer 1

The correct option is A (a)$x^2+y^2=1$

Let P($x_1,y_1)$ be the external point then the equation of chord of contact of circle $x^2+y^2=1$ is given as

$x{x}_1+y{y}_1=1$

Since, above line touches the circle $x^2+y^2=4$

hence chord of contact will satisfy the equation of circle at a single point as follows

$x^2+(\frac{1-x{x}_1}{y_1}-1{)}^2=1$

$(x_1^2+y_1^2)x_2+2x_1(y_1-1)x+(1-2y_1)=0$

Since, the lines the circle at a single point hence above quadratic equation must have equal roots

i.e. its determinant $b^2-4c$ must be zero. Hence we have

$\left(2x_1\right(y_1-1){)}^2-4(x_1^2+y_1^2\left)\right((1-2y_1))=0$

$4x_1^2{y}_1^2+4x_1^2-8x_1^2{y}_1-4x_1^2+8x_1^2{y}_1-4y_1^2+8y_1^3=0$

$x_1^2+2y_1-1=0$

setting $x_1=x,y_1=y$ in above equation the locus of the parametric point $P(x_1,y_1)$ is given as

$x^2+2y-1=0$