AP EAMCET · Maths · Circle
If \((a, b)\) is the common point for the circles \(x^2+y^2-4 x+4 y-1=0\) and \(x^2+y^2+2 x-4 y+1=0\), then \(a^2+b^2=\)
- A \(\frac{1}{5}\)
- B 5
- C 25
- D \(\frac{1}{25}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{5}\)
Step-by-step Solution
Detailed explanation
\((x^2+y^2-4 x+4 y-1) - (x^2+y^2+2 x-4 y+1) = 0 \Rightarrow -6x+8y-2=0 \Rightarrow 3x-4y+1=0\) For common point \((a,b)\), \(3a-4b+1=0 \Rightarrow 4b=3a+1\). Also, \(a^2+b^2-4a+4b-1=0\). Substitute \(4b\): \(a^2+b^2-4a+(3a+1)-1=0 \Rightarrow a^2+b^2=a\). Substitute…
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