A) \[{{(x-\sqrt{7})}^{2}}+{{(y-1)}^{2}}=1\]
B) \[{{(x+\sqrt{7})}^{2}}+{{(y-1)}^{2}}=1\]
C) \[{{x}^{2}}+{{y}^{2}}=2\sqrt{7}x+2y\]
D) None of these
Correct Answer: A
Solution :
[a] We have \[{{x}^{2}}+{{y}^{2}}+4y-5=0\], Its centre is \[{{C}_{1}}(0,-2),\] |
\[{{r}_{1}}=\sqrt{4+5}=3.\] Let \[{{C}_{2}}(h,k)\] be the centre of the smaller circle and its radius \[{{r}_{2}}.\] Then \[{{C}_{1}}{{C}_{2}}=4.\] |
\[\Rightarrow \sqrt{{{h}^{2}}+{{(k+2)}^{2}}}=3+{{r}_{2}}=4\] ?. (1) |
\[\Rightarrow {{r}_{2}}=1\] |
But \[k={{r}_{2}}=1\] |
[it touches x-axis] |
\[\therefore \] From eq. (1), \[4=\sqrt{{{h}^{2}}+{{(1+2)}^{2}}}\] |
\[\Rightarrow \,\,\,\,\,\,16={{h}^{2}}+9\Rightarrow {{h}^{2}}=7\Rightarrow h=\pm \sqrt{7}\] |
Since \[h>0\therefore h=\sqrt{7}\] |
Hence, required circle is |
\[{{(x-\sqrt{7})}^{2}}+{{(y-1)}^{2}}=1\] |
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