WBJEE · Maths · Circle
The equation of circle of radius \(\sqrt{17}\) unit, with centre on the positive side of \(x\) -axis and through the point \((0,1)\) is
- A \(x^{2}+y^{2}-8 x-1=0\)
- B \(x^{2}+y^{2}+8 x-1=0\)
- C \(x^{2}+y^{2}-9 y+1=0\)
- D \(2 x^{2}+2 y^{2}-3 x+2 y=4\)
Answer & Solution
Correct Answer
(A) \(x^{2}+y^{2}-8 x-1=0\)
Step-by-step Solution
Detailed explanation
Hint: Let centre be \(\equiv(a, 0)(a>0)\) So, circle : \((x-a)^{2}+y^{2}=17 ;\) as it passes through \((0,1)\), so \(a^{2}+1=17 \quad \Rightarrow a=4(a \neq-4, a>0)\) \(\therefore\) Equation is : \(x^{2}+y^{2}-8 x-1=0\)
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