Question

The equation of one of the tangents from (1, 1) to a circle with its centre at (3, 0) is 3x + y – 4 = 0. The equation of the other tangent is-

• A. 5x – y – 4 = 0
• B. 3y – x – 2 = 0
• C. 3y + x – 4 = 0
• D. 3x – y – 2 = 0

Answer 1

Answer: (B)

3y – x – 2 = 0

Answer 2

If AB is the line 3x + y – 4 = 0 and C(3, 0)

BC = $\sqrt { \frac { 5 } { 2 } }$, AC =$\sqrt { 5 }$

\ q = sin–1 $\frac { \pi } { 4 }$

Hence the two tangents are at right angles. The other tangent is y – 1 = $\frac { 1 } { 3 }$ (x – 1)

or 3y – x – 2 = 0.