AP EAMCET · Maths · Circle
Tangent \(L_1 \equiv 3 x-4 y-8=0\) and the chord \(L_2 \equiv x+y-1=0\) are at a distance of 2 and \(\sqrt{2}\) units respectively from the centre of a circle S. \((h, k)\) is the centre of S such that \(h^2+\) \(\mathrm{k}^2=13\). If the midpoint of the chord \(\mathrm{L}_2=0\) is \((\alpha, \beta)\) and the radius of the circle is \(\mathrm{r}\), then \(\alpha+\beta+\mathrm{r}=\)
- A \(4\)
- B \(-1\)
- C \(7\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
Radius of circle \(=2=r\) \(\because(\alpha, \beta)\) passes through \(\mathrm{L}_2\). \(\begin{aligned} & \therefore \alpha+\beta-1=0 \Rightarrow \alpha+\beta=1 \\ & \text { Now } \alpha+\beta+r=2+1=3 \end{aligned}\)
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