A) \[-\frac{1}{2}\le k\le \frac{1}{2}\]
B) \[k\le \frac{1}{2}\]
C) \[0\le k\le \frac{1}{2}\]
D) \[k\ge \frac{1}{2}\]
Correct Answer: D
Solution :
Equation of circle whose centre is (h, k) is given as \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{k}^{2}}\] |
(radius of circle = k because circle is tangent to x-axis) |
Equation of circle passing through \[(-1,+1)\] |
\[\therefore \,\,{{(-1-h)}^{2}}+{{(1-k)}^{2}}={{k}^{2}}\] |
\[\Rightarrow \,\,1+{{h}^{2}}+2h+1+{{k}^{2}}-2k={{k}^{2}}\] |
\[\Rightarrow \,\,{{h}^{2}}+2h-2k+2=0\] |
Since, it is a quadratic in 'h' and \[D\ge 0\] |
\[\therefore \,\,{{(2)}^{2}}-4\times 1.(-2k+2)\ge 0\] |
\[\Rightarrow \,\,4-4(-2k+2)\ge 0\Rightarrow 1+2k-2\ge 0\Rightarrow k\ge \frac{1}{2}\] |
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