First slide

Analysis of two circles in circles

Question

The locus of the center 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

$\begin{array}{rcl} Let the circle be x^{2} + y^{2} + 2 g x + 2 f y + c & = & 0 \dots \dots (1) \\ It cuts x^{2} + y^{2} - 20 x + 4 & = & 0 orthogonally \\ \begin{aligned} 2 [- 10 g + 0 x f] = c + f \\ \Rightarrow - 20 g = c + 4 \dots \dots \dots \dots \dots (2) \end{aligned} \\ Circle (1) touches line x & = & 2 \\ \begin{aligned} |\frac{- g - 2}{\sqrt{1^{2} + 0}}| = \sqrt{g^{2} + f^{2} - c} (\sin c e d = r) \\ \Rightarrow (g + 2)^{2} = g^{2} + f^{2} - c \\ 4 g + 4 = f^{2} - c \dots \dots \dots (3) \end{aligned} \\ From (\underline{2)} & (3) eliminate ' c' \\ \begin{aligned} - 16 g + 4 = f^{2} + 4 \\ \Rightarrow f^{2} + 16 g = 0 \end{aligned} \\ L o c u s o f (- g, - f) i s \\ \begin{aligned} y^{2} - 16 x = 0 \\ y^{2} = 16 x \end{aligned} \end{array}$

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