A) \[x=0,y=0\]
B) \[x=1,y=0\]
C) \[({{h}^{2}}-{{r}^{2}}))x-2rhy=0,y=0\]
D) \[({{h}^{2}}-{{r}^{2}})x-2rhy=0,x=0\]
Correct Answer: D
Solution :
Since the given equation of the circle is \[{{x}^{2}}+{{y}^{2}}-2rx-2hy+{{h}^{2}}=0\]whose centre is\[(r,h)\]nd radius is r. \[\therefore \]\[x=0\]is one of the tangent. Let another tangent is\[y=mx\]to circle. This line will be tangent, if \[\frac{h-mr}{\sqrt{1+{{m}^{2}}}}=r\] \[\Rightarrow \] \[{{h}^{2}}+{{m}^{2}}{{r}^{2}}-2hmr={{r}^{2}}+{{r}^{2}}{{m}^{2}}\] \[\Rightarrow \] \[m=\frac{({{h}^{2}}-{{r}^{2}})}{2hr}\] \[\therefore \]Equation of tangent is \[y=\frac{{{h}^{2}}-{{r}^{2}}}{2hr}x\] \[\Rightarrow \]\[2hry=({{h}^{2}}-{{r}^{2}})x\]and\[x=0\]are two tangents.
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