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 \[\Rightarrow ({{x}^{2}}+{{y}^{2}})-2(hx+ky)\left( \frac{x}{2h}+\frac{y}{2k} \right)+\]You need to login to perform this action.
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