Question

The locus of the centre of a circle which touches the circle

x² + y² – 4x – 4y + 7 = 0 externally and also touches the Y-axis is given by the equation –

• A. x² + 5x – 3y + 1 = 0
• B. x² – 7x + 6y – 5 = 0
• C. y² – 4y – 6x + 7 = 0
• D. y² – 7y – 6x + 3 = 0

Answer 1

Answer: (C)

y² – 4y – 6x + 7 = 0

Answer 2

The given circle has centre at C ŗ (2, 2) and radius

r = $\sqrt{2^{2} + 2^{2} - 7}$ = 1

Let P(h, k) be the centre and |h| be the radius of the variable circle, then we have CP = |h| + 1

i.e.(h – 2)² + (k – 2)² = (|h| + 1)²

i.e. h² + k² – 4h – 4k + 8 = h² + 2|h| + 1

Since the variable circle can lie in the 1^st or the 4^th quadrant only, therefore |h| = h

The above equation thus reduces to k² – 4k – 6k + 7 = 0

Hence, the locus of P is y² – 4y – 6x + 7 = 0.