Given that a > 2b > 0 and that the line y = mx - b \sqrt{1+m^2} is a common tangent to the circles x^2+y^2 = b^2 and (x-a)^2+y^2 = b^2. Then the positive value of m is

Correct Answer is : 2b√a2−4b2

AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

Section: Mathematics

Question : 50 of 160

Marks: +1, -0

Given that a > 2b > 0 and that the line y = mx - b √1+m2 is a common tangent to the circles x2+y2 = b2 and (x−a)2+y2 = b2. Then the positive value of m is

2b
a−2b
b
a−2b
√a2−4b2
2b
2b
√a2−4b2

Validate

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