COMEDK · Maths · 12. Circle
The centre of the circle passing through \((0,0)\) and \((1,0)\) and touching the circle \(x^2+y^2=9\) is
- A \(\left(\dfrac{1}{2}, \dfrac{3}{2}\right)\)
- B \(\left(\dfrac{3}{2}, \dfrac{1}{2}\right)\)
- C \(\left(\dfrac{1}{2},-\sqrt{2}\right)\)
- D \(\left(\dfrac{1}{2}, \dfrac{1}{2}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\dfrac{1}{2},-\sqrt{2}\right)\)
Step-by-step Solution
Detailed explanation
Centre lies on perpendicular bisector of \((0,0)\) and \((1,0)\), which is \(x = \dfrac{1}{2}\). So \(h = \dfrac{1}{2}\). \(r^2 = h^2 + k^2 = \dfrac{1}{4} + k^2\) Distance between centres \(C_1 = \left(\dfrac{1}{2}, k\right)\) and \(C_2 = (0,0)\):…
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