AP EAMCET · Maths · Circle
Let the circle \(\mathrm{S}\) which is concentric with the circle \(x^2+y^2-2 x+k y+4=0\) pass through the point \((3,-2)\). If one of the diameters of S lies along the line \(3 x-2 y+\) \(4=0\), then the radius of the circle \(S\) is
- A \(\frac{\sqrt{149}}{2}\)
- B \(\sqrt{31}\)
- C \(\sqrt{38}\)
- D \(\frac{1}{2} \sqrt{137}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{2} \sqrt{137}\)
Step-by-step Solution
Detailed explanation
The given equation of circle is \(x^2+y^2-2 x+k y+4=0\) ...(i) Centre of above circle (i) is \(c \equiv\left(1,-\frac{k}{2}\right)\) \(\because\) Circle \(S\) is concentric with circle (i) \(\therefore \quad\) Equation of \(S\) is \((x-1)^2+\left(y+\frac{k}{2}\right)^2=r^2\)…
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