AP EAMCET · Maths · Circle
Let \(P\) and \(Q\) be the inverse points with respect to the circle \(S \equiv x^2+y^2-4 x-6 y+k=0\) and \(C\) be the centre of the circle \(S=0\) such that \(C P\). \(C Q=4\). If \(P=(1,2)\) and \(\mathrm{Q}=(\mathrm{a}, \mathrm{b})\), then \(2 \mathrm{a}=\)
- A b
- B \(-1\)
- C \(3 \mathrm{~b}\)
- D \(0\)
Answer & Solution
Correct Answer
(D) \(0\)
Step-by-step Solution
Detailed explanation
We have \(C P \cdot C Q=r^2=4 \Rightarrow r=2\) Centre of given circle \(x^2+y^2-4 x-6 y+k=0\) \(=C(2,3)\). We know that inverse of point \(P(\alpha, \beta)\) with respect to circle with centre \((h, k)\) and radius \(r\) is the point \((a, b)\) then \(a=\lambda(\alpha-h)+h\)…
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