Question

The locus of the centre of a circle which cuts orthogonally the circle $x^{2} + y^{2} - 20x + 4 = 0$ and which touches $x = 2$ is

• A. $y^{2} = 16x + 4$
• B. $x^{2} = 16y$
• C. $x^{2} = 16y + 4$
• D. $y^{2} = 16x$

Answer 1

Answer: (D)

$y^{2} = 16x$

Answer 2

Let the circle be $x^{2} + y^{2} + 2gx + 2fy + c = 0$ …..(i)

It cuts the circle $x^{2} + y^{2} - 20x + 4 = 0$ orthogonally

$\therefore$ $2( - 10g + 0 \times f) = c + 4$

$\Rightarrow - 20g = c + 4$ …..(ii)

Circle (i) touches the line $x = 2;$

$\therefore$ $x + 0y - 2 = 0$

$\therefore$ $\left| \frac{- g + 0 - 2}{\sqrt{1}} \right| = \sqrt{g^{2} + f^{2} - c}$

$\Rightarrow (g + 2)^{2} = g^{2} + f^{2} - c$ $\Rightarrow 4g + 4 = f^{2} - c$ …..(iii)

Eliminating c from (ii) and (iii), we get

$- 16g + 4 = f^{2} + 4 \Rightarrow f^{2} + 16g = 0$.

Hence the locus of ($- g, - f)$ is $y^{2} - 16x = 0$

$\Rightarrow y^{2} = 16x.$