Question

List- I	List- II
A. The locus of the point of intersection of the perpendicular tangents of $x^2+y^2=a^2$	1)$x^2+y^2=5$ of
B. The locus of the mid-point of the chord of the circle $x^2+y^2=8$ which is $\sqrt{2}$ units from origin is	2)$x^2+y^2=a^2{\sec }^2\alpha$
C. The locus of the point of intersection of the lines $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2t}{1+t^2}$	3)$x^2+y^2=2a^2$
D. The locus of the point whose chord of contact w.r.t $x^2+y^2=a^2$ making an angle '2$\alpha$' at the center is	4)$x^2+y^2=2$ 5)$x^2+y^2=1$

• A. $3452$
• B. $3541$
• C. $2413$
• D. $2453$

Answer 1

Answer: (A)

$3452$

(A) P is a locus of point of intersation of perperdiculer tangents.
$\therefore OA=AP=r=a$
$\therefore OP=\sqrt{2}r.$ Using Pythegores theorem.
$\therefore h^2+k^2=2r^2=2a^2$
$\therefore x^2+y^2=2a^2$ is a locus of $P(h,k)$.

(B) Let P$(h,k)$ is a middle point of chord , which is $\sqrt{2}$ units from origin
$hx+ky-8=h^2+k^2-8$
$hx+ky=h^2+k^2$
$\therefore \sqrt{2}=\left|\frac{h^2+k^2}{\sqrt{h^2+k^2}}\right|$
$2=h^2+k^2$

So locus of P(h,k)is $x^2+y^2=2$

(C) Given $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2t}{1+t^2}$

Upon squaring and adding we get $x^2+y^2=1$

(D) Locus of p(h,k)

$\tan \alpha =\frac{PA}{OA}=\frac{\sqrt{h^2+k^2-a^2}}{a}$

$h^2+k^2=a^2+ta{n}^2\alpha a^2$
$h^2+k^2=a^{2}se{c}^2\alpha$
locus of p(h,k) is $x^2+y^2=a^{2}se{c}^2\alpha$