A) \[c=h\pm k\]
B) \[{{c}^{2}}={{h}^{2}}+{{k}^{2}}\]
C) \[{{c}^{2}}={{(h+k)}^{2}}\]
D) \[4{{c}^{2}}={{h}^{2}}+{{k}^{2}}\]
Correct Answer: B
Solution :
The line is \[\frac{x}{2h}+\frac{y}{2k}=1\]and circle is, \[{{x}^{2}}+{{y}^{2}}-2(hx+ky)+({{h}^{2}}+{{k}^{2}}-{{c}^{2}})=0\] Making it homogeneous, we get \[\Rightarrow ({{x}^{2}}+{{y}^{2}})-2(hx+ky)\left( \frac{x}{2h}+\frac{y}{2k} \right)+\] If these lines be perpendicular, then \[A+B=0\] \[\left[ 1-1+\frac{({{h}^{2}}+{{k}^{2}}-{{c}^{2}})}{4{{h}^{2}}} \right]+\left[ 1-1+\frac{({{h}^{2}}+{{k}^{2}}-{{c}^{2}})}{4{{k}^{2}}} \right]=0\] or \[({{h}^{2}}+{{k}^{2}}-{{c}^{2}})\text{ }\left( \frac{{{h}^{2}}+{{k}^{2}}}{4{{h}^{2}}{{k}^{2}}} \right)=0\] \[\therefore {{h}^{2}}+{{k}^{2}}={{c}^{2}}\].You need to login to perform this action.
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