Question

Let $L_1,L_2$ and $L_3$ be the lengths of tangents drawn from a point $P(h,k)$ to the circles $x^2+y^2=4,$ $x^2+y^2–4x=0$ and $x^2+y^2–4y=0$ respectively. Let $L_1^4=L_2^2{L}_3^2+16$ gives the locus of $P$ as two curves, $C_1$ (straight line) and $C_2$ (circle). A triangle is formed by $C_1$ and two other lines which are at an angle of ${45}^{\circ }$ with $C_1$ and tangent to the circle $C_2.$ Then which of the following is/are correct?

• A. Straight line which intersects both the curves $C_1$ and $C_2$ orthogonally, is $x–y=0$.
• B. Straight line which intersects both the curves $C_1$ and $C_2$ orthogonally, is $x+y=0$.
• C. Circumcenter of the triangle is at origin.
• D. Circumcenter of the triangle is at $(1,1).$

Answer 1

Answer: (A), (C)

The correct options are
A Straight line which intersects both the curves $C_1$ and $C_2$ orthogonally, is $x–y=0$.
C Circumcenter of the triangle is at origin.

Clearly,
$L_1=\sqrt{h^2+k^2-4},L_2=\sqrt{h^2+k^2-4h},L_3=\sqrt{h^2+k^2-4k}$

Now, $L_1^4=L_2^2{L}_3^2+16$ (Given)
$\Rightarrow (h^2+k^2-4{)}^2=(h^2+k^2-4h\left)\right(h^2+k^2-4k)+16$
$\Rightarrow (h^2+k^2{)}^2-8(h^2+k^2)+16=(h^2+k^2{)}^2-4(h^2+k^2)(h+k)+16hk+16$
$\Rightarrow 8(h+k{)}^2-4(h^2+k^2\left)\right(h+k)=0$
$\Rightarrow 4(h+k)[h^2+k^2-2(h+k\left)\right]=0$
So, locus of $P(h,k)$ are
$C_1:x+y=0$ and $C_2:x^2+y^2-2x-2y=0$

From the figure, it is clear that straight line which intersects both the curves orthogonally is, $x–y=0$.


$\Delta PQR$ is required triangle which is right angled isosceles triangle and its circumcenter is $(0,0)$.