Question

The locus of the centre of a circle touching the circle

x² + y² – 4y – 2x = 2$\sqrt { 3 }$ – 1 internally and tangents on which from (1, 2) is making a 60ŗ angle with each other is –

• A. (x – 1)² + (y – 2)² = 3
• B. (x – 2)² + (y – 1)² = 1 + 2$\sqrt { 3 }$
• C. x² + y² = 1
• D. None of these

Answer 1

Answer: (A)

(x – 1)² + (y – 2)² = 3

Answer 2

Let r & R be the radii of required and given circles resp. & let centre is (h, k).

By given condition

= R – r.

Now = tan 30⁰

Ž r = AB tan 30⁰

= (R – r) $\frac { 1 } { \sqrt { 3 } }$ (AB = R – r)

Ž == R – $\frac { \mathrm { R } } { 1 + \sqrt { 3 } }$ = R $\left( \frac { \sqrt { 3 } } { 1 + \sqrt { 3 } } \right)$

Now R = 1 +$\sqrt { 3 }$

\ Locus is ( x – 1)² + (y – 2)² = 3