AP EAMCET · Maths · Circle
The radical centre of the circles \(x^2+y^2-4 x-6 y+5=0\), \(x^2+y^2-2 x-4 y-1=0\) and \(x^2+y^2-6 x-2 y=0\) is equal to
- A \(\left(\frac{33}{4}, \frac{20}{3}\right)\)
- B \(\left(\frac{33}{4}, \frac{10}{3}\right)\)
- C \(\left(\frac{33}{4}, \frac{-20}{3}\right)\)
- D \(\left(\frac{7}{6}, \frac{11}{6}\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{7}{6}, \frac{11}{6}\right)\)
Step-by-step Solution
Detailed explanation
Equation of given circles \(\begin{aligned} & S_1: x^2+y^2-4 x-6 y+5=0 \\ & S_2: x^2+y^2-2 x-4 y-1=0 \end{aligned}\) and \(\quad S_3: x^2+y^2-6 x-2 y=0\) \(\therefore\) Radical axis of circles \(S_1\) and \(S_2\) is \(2 x+2 y-6=0 \Rightarrow x+y=3\)...(i) Similarly, the radical…
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