AP EAMCET · Maths · Circle
The slope of one of the direct common tangents drawn to the circles \(x^2+y^2-2 x+4 y+1=0\) and \(x^2+y^2-4 x-2 y+4=0\) is
- A \(0\)
- B \(\frac{4}{3}\)
- C \(\frac{3}{4}\)
- D \(1\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{3}\)
Step-by-step Solution
Detailed explanation
\(C_1 = (1, -2), R_1 = \sqrt{1^2+(-2)^2-1} = 2\) \(C_2 = (2, 1), R_2 = \sqrt{2^2+1^2-4} = 1\) Let the tangent be \(mx-y+c=0\). For direct common tangents: \(\frac{m(1)-(-2)+c}{2} = \frac{m(2)-(1)+c}{1}\) \(m+2+c = 2(2m-1+c)\) \(m+2+c = 4m-2+2c\) \(c = 4-3m\) Distance from…
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