AP EAMCET · Maths · Circle
\(\mathrm{P}\) is a point of intersection of the circles \(\mathrm{S} \equiv \mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}\) \(+2 k y+1=0\) and \(S^1 \equiv x^2+y^2+2 k x-6 y-7=0\). If the tangent at \(P\) to \(S=0\) pass through the centre of \(S^1=0\) and the tangent at \(\mathrm{P}\) to \(\mathrm{S}^1=0\) pass through the centre of \(\mathrm{S}=0\), then the radius of \(\mathrm{S}^1=0\) is
- A \(\frac{\sqrt{33}}{2}\)
- B \(33\)
- C \(\sqrt{17}\)
- D \(\frac{\sqrt{65}}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\sqrt{65}}{2}\)
Step-by-step Solution
Detailed explanation
From given equation of circles \(\begin{aligned} & \mathrm{C}^{\prime}(3,-\mathrm{k}), \mathrm{r}^{\prime}=\sqrt{9+\mathrm{k}^2-1} \\ & \text { and } \mathrm{C}^{\prime}(-\mathrm{k}, 3), \mathrm{r}^{\prime}=\sqrt{9+\mathrm{k}^2+7} \end{aligned}\) According to question…
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