First slide

Definition of a circle

Question

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

Moderate

A

$y^{2} = 16 x + 4$
B

$x^{2} = 16 y$
C

$x^{2} = 16 y + 4$
D

$y^{2} = 16 x$

Solution

Let the circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
It cuts the circle $x^{2} + y^{2} - 20 x + 4 = 0$ orthogonally
$\begin{array}{ll} ∴ 2 (- 10 g + 0 \times f) = c + 4 \\ \Rightarrow - 20 g = c + 4 \end{array}$
Circle (i) touches the line $x = 2$ i.e. x + $O y -$ $2 = 0$
$\begin{aligned} ∴ |\frac{- g + 0 - 2}{\sqrt{1^{2} + 0^{2}}}| = \sqrt{g^{2} + f^{2} - c} \\ \Rightarrow (g + 2)^{2} = g^{2} + f^{2} - c \Rightarrow 4 g + 4 = f^{2} - c \end{aligned}$
Eliminating c from (ii) and (iii), we get
$- 16 g + 4 = f^{2} + 4 \Rightarrow f^{2} + 16 g = 0$
Hence, the locus of $(- g, - f) is y^{2} - 16 x = 0$

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