AP EAMCET · Maths · Circle
A Circle S passes through the points of intersection of the circles \(x^2+y^2-2 x+2 y-2=0\) and \(x^2+y^2+2 x-2 y+1=0\). If the centre of this circle S lies on the line \(x-y+6=0\), then the radius of the circle \(S\) is
- A \(\sqrt{5}\)
- B \(5\)
- C \(\sqrt{41}\)
- D \(\sqrt{14}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{14}\)
Step-by-step Solution
Detailed explanation
Since equation of circles is \(x^2+y^2-2 x+2 y-2+\) \(\begin{aligned}& k\left(x^2+y^2+2 x-2 y+1\right)=0\\&\Rightarrow(1+k) x^2+(1+k) y^2+(2 k-2) x+(2-2 k) y+k-2=0\end{aligned}\) \(\Rightarrow x^2+y^2+\frac{2(k-1)}{k+1} x+\frac{2(1-k)}{1+k} y+\frac{k-2}{k+1}=0\) So, centre…
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