Question

If two circles, each of radius 5 units, touch each other at (1, 2) and the equation of their common tangent is 4x + 3y = 10, then equation of the circle, a portion of which lies in all the quadrants is –

• A. x² + y² – 10x – 10y + 25 = 0
• B. x² + y² + 6x + 2y – 15 = 0
• C. x² + y² + 2x + 6y – 15 = 0
• D. x² + y² + 10x + 10y + 25 = 0

Answer 1

Answer: (B)

x² + y² + 6x + 2y – 15 = 0

Answer 2

The centres of the two circles will lie on the line through P(1, 2) perpendicular to the common tangent 4x + 3y = 10. If C₁ and C₂ are the centres of these circles then PC₁ = 5 = r₁, PC₂ = –5 = r₂. Also C₁, C₂ lie on the line

$\frac{x - 1}{\cos\theta}$ = $\frac{y - 2}{\sin\theta}$ = r where tan q = 3/4.

When r = r₁ the coordinates of C₁ are (5 cos q + 1, 5 sin q + 2) or (5, 5) as cos q = 4/5, sin q = 3/5.

When r = r₂, the coordinates of C₂ are (–3, –1)

The circle with centre C₁(5, 5) and radius 5 touches both the coordinate axes and hence lies completely in the first quadrant.

Therefore the required circle is with centre

(–3, –1) and radius 5, so its equation is

(x + 3)² + (y + 1)² = 5² or x² + y² + 6x + 2y – 15 = 0

Since the origin lies inside the circle, a portion of the circle lies in all the quadrants.