AP EAMCET · Maths · Circle
The equation to the line joining the centres of the circles belonging to the coaxial system of circles
\(4 x^2+4 y^2-12 x+6 y-3+\lambda(x+2 y-6)=0\)
is
- A \(8 x-4 y-15=0\)
- B \(8 x-4 y+15=0\)
- C \(3 x-4 y-5=0\)
- D \(3 x-4 y+5=0\)
Answer & Solution
Correct Answer
(A) \(8 x-4 y-15=0\)
Step-by-step Solution
Detailed explanation
Given coaxial system of circle is \(\begin{aligned} & 4 x^2+4 y^2-12 x+6 y-3+\lambda(x+2 y-6)=0 \\ & \text { or } x^2+y^2-3 x+\frac{3 y}{2}-\frac{3}{4}+\frac{\lambda}{4}(x+2 y-6)=0\end{aligned}\) Here, radical axis is \(x+2 y-6=0\). i.e., line of centre is \(2 x-y+k=0\) Here,…
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