Question

The locus of the centre of a circle which cuts orthogonally the circle x² – 20x + y² + 4 = 0 and which touches x = 2 is

• A. y² = 16x + 4
• B. x² = 16y
• C. x² = 16 y + 4
• D. y² = 16x

Answer 1

Answer: (D)

y² = 16x

Answer 2

Let the general equation of circle be

x² + y² + 2gx + 2fy + c = 0 ……. (i)

its cuts the circle x² + y² – 20x + 4 = 0 orthogonally

\ 2 (–10g + 0 ×f ) = c + 4 Ž – 20 g = c + 4 …….(ii)

Q circle (i) touches x = 2

\ perpendicular distance from centre to the circle = radius

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

Ž (g + 2)² = g² + f² –c

Ž g² + 4 + 4g = g² + f² –c Ž 4g + 4 = f ² –c … (iii)

on eliminating c form (ii) and (iii) we get

–16g + 4 = f ² + 4 Ž f ² + 16g = 0

Hence locus of (–g, –f) is

y² – 16x = 0 (replacing –f & –g by x & y)