TS EAMCET · Maths · Circle
The length of the common chord of the two circles \(x^2+y^2-4 y=0\) and \(x^2+y^2-8 x\) \(-4 y+11=0\), is
- A \(\frac{\sqrt{145}}{4} \mathrm{~cm}\)
- B \(\frac{\sqrt{11}}{2} \mathrm{~cm}\)
- C \(\sqrt{135} \mathrm{~cm}\)
- D \(\frac{\sqrt{135}}{4} \mathrm{~cm}\)
Answer & Solution
Correct Answer
(D) \(\frac{\sqrt{135}}{4} \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
Given equation of circles are and \[ \begin{gathered} x^2+y^2-4 y=0 \\ x^2+y^2-8 x-4 y+11=0 \end{gathered} \] \(\therefore\) Equation of chord \[ \begin{aligned} & \quad x^2+y^2-4 y-\left(x^2+y^2-8 x-4 y+11\right)=0 \\ & \Rightarrow \quad 8 x-11=0 \end{aligned} \] Centre and…
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