A) \[\{(x,y):{{x}^{2}}=4y\}\cup \{(x,y):y\le 0\}\]
B) \[\{(x,y):{{x}^{2}}+{{(y-1)}^{2}}=4\}\cup \{(x,y):y\le 0\}\]
C) \[\{(x,y):{{x}^{2}}=y\}\cup \{(0,y):y\le 0\}\]
D) \[\{(x,y):{{x}^{2}}=4y\}\cup \{(0,y):y\le 0\}\]
Correct Answer: D
Solution :
Let the centre of circle C be (h, k). Then as this circle touches axis of\[x\], its radius \[=\left| k \right|\] |
Also it touches the given circle \[{{x}^{2}}+{{(y-1)}^{2}}=1,\] |
Centre (0, 1) radius 1, externally |
Therefore, the distance between centres = sum of radii |
\[\Rightarrow \]\[\sqrt{{{(h-0)}^{2}}+{{(k-1)}^{2}}}=1+\left| k \right|\] |
\[\Rightarrow \]\[{{h}^{2}}+{{k}^{2}}-2k+1={{(1+\left| k \right|)}^{2}}\] |
\[\Rightarrow \]\[{{h}^{2}}+{{k}^{2}}-2k+1=1+2\left| k \right|+{{k}^{2}}\] |
\[\Rightarrow \]\[{{h}^{2}}=2k+2\left| k \right|\] |
\[\therefore \]locus of \[(h,k)\] is, \[{{x}^{2}}=2y+2\left| y \right|\] |
Now if \[y>0,\]it becomes\[{{x}^{2}}=4y\] |
and if \[y\le 0,\]it becomes \[x=0\] |
\[\therefore \]Combining the two, the required locus is |
\[\{(x,y):{{x}^{2}}=4y\}\cup \{(0,y):y\le 0\}\] |
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