AP EAMCET · Maths · Ellipse
The lines \(y=2 x+\sqrt{76}\) and \(2 y+x=8\) touch the ellipse \(\frac{x^2}{16}+\frac{y^2}{12}=1\). If the point of intersection of these two lines lie on a circle. whose centre coincides with the centre of that ellipse, then the equation of that circle is
- A \(x^2+y^2=28\)
- B \(x^2+y^2=16\)
- C \(x^2+y^2=12\)
- D \(x^2+y^2=(4+\sqrt{8})^2\)
Answer & Solution
Correct Answer
(A) \(x^2+y^2=28\)
Step-by-step Solution
Detailed explanation
Given, equation of tangents to the ellipse are, \[ \begin{aligned} & y=2 x+\sqrt{76} \text { and } 2 y+x=8 \Rightarrow y=-\frac{1}{2} x+4 \\ & \Rightarrow \quad m_1=2 \text { and } m_2=-- \\ & \because \quad \text { Here, } m_1 \times m_2=-1 \end{aligned} \] It means that the…
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