question_answer

If the equation of the common tangent at the point \[(1,-1)\] to the two circles, each of radius 13, is \[12x+5y-7=0\] then the centres of the two circles are

A) \[(13,4),(-11,6)\]

B) \[(13,4),(-11,-6)\]

C) \[(13,-4),(-11,-6)\]

D) \[(-13,4),(-11,-6)\]

Correct Answer: B

Solution :

[b] Let A, B, be the centres of the two circles, Slope of the common tangent \[=-\frac{12}{5}\] \[\therefore \] Slope of AB is \[\tan \theta =-\frac{1}{-\frac{12}{5}}=\frac{5}{12}\] The point (1, -1) lies on the line AB and the points A and B are at a distance 13 from the point (1,-1) \[\therefore \] Coordinates of A and B are \[(1\pm 13cos\theta ,-1\pm 13\sin \theta ),\] where \[\tan \theta =\frac{5}{12}\] i.e. \[\left( 1\pm 13\frac{12}{13},-1\pm 13\frac{5}{13} \right)\] or \[(1\pm 12,-1\pm 5)\] i.e., \[(13,4)\] and \[(-11,-6)\]